The mathematical constant
is a transcendental number (or inalgebraic).
In other words, it is not a root of any polynomial with rational coefficients.
Let us assume that
is algebraic, so there exists a polynomial
![{\displaystyle P(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{d}z^{d}\in \mathbb {Q} [z],\quad (a_{0}\neq 0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff89530e7782aa30a7b4ccd5cffb97c7f7cbc9dc)
such that
.
Lemma: If
is algebraic, then
is algebraic.
Proof: We get
![{\displaystyle {\begin{aligned}P(\pm iz)&=a_{0}+a_{1}(\pm iz)+a_{2}(\pm iz)^{2}+\cdots +a_{d}(\pm iz)^{d}\\[5pt]&=(a_{0}-a_{2}z^{2}+\cdots \,)\pm (a_{1}z-a_{3}z^{3}+\cdots \,)\,i\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34c8f2c646bbd6082bd7deff29bd057969fddf55)
hence
is a root of the polynomial
![{\displaystyle P(iz)P(-iz)=(a_{0}-a_{2}z^{2}+\cdots \,)^{2}+(a_{1}z-a_{3}z^{3}+\cdots \,)^{2}\in \mathbb {Q} [z]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53bac4a1d308719ec27a8a12cc682673e4afe23f)
Therefore, there exists a polynomial
of degree
with roots
, such that
.
By Euler's identity we have
. Therefore:
![{\displaystyle {\begin{aligned}0&=({\text{e}}^{z_{1}}\!+1)({\text{e}}^{z_{2}}\!+1)\cdots ({\text{e}}^{z_{n}}\!+1)\\[3pt]&=\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}+\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}{\text{e}}^{z_{i_{3}}}\!+\cdots +\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<\cdots <i_{n}\leq n}\!\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}\!\cdots {\text{e}}^{z_{i_{n}}}\!+1\\[3pt]&=\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}}+\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}+\,z_{i_{3}}}\!+\cdots +{\text{e}}^{z_{1}+\,\cdots \,+\,z_{n}}\!+{\text{e}}^{0}\\[3pt]&=\sum _{i\,=\,1}^{2^{n}}{\text{e}}^{\beta _{i}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc54c09520f8405e96f5e10b955b992f45504264)
The exponents are symmetric polynomial in
, and among them are
non-zero sums. That is:

As we previously learned, for all
there exists a monic polynomial
of degree
such that its roots are the sums of every
of the roots
. Therefore:
![{\displaystyle {\begin{aligned}Q(z)&=P_{1}(z)\cdots P_{n}(z)\in \mathbb {Q} [z]\\[5pt]&=(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\cdots (z-\beta _{2^{n}})\\[5pt]&=z^{2^{n}-\,m}(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5117cfb83e5cff87d5e14cb0e96cee9d13dbe0a6)
After reduction we get that:
![{\displaystyle (z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\in \mathbb {Q} [z]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc30207560d8c005a5a36375bd5f66448d0d203c)
Multiplying by the least common multiple
of the rational coefficients, we get a polynomial of the form
![{\displaystyle {\begin{aligned}B(z)&=b_{m}(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\\[5pt]&=b_{m}z^{m}+b_{m-1}z^{m-1}+\cdots +b_{1}z+b_{0}\in \mathbb {Z} [z]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6db3f5aaef8539151a09eb358878c9f03d228b93)
Let
be a polynomial of degree
. Let us define
. Taking its derivative yields:

Let us define
. Taking its derivative yields:
![{\displaystyle G'\!(z)={\text{e}}^{-z}F'\!(z)-{\text{e}}^{-z}F(z)={\text{e}}^{-z}{\bigl [}F'\!(z)-F(z){\bigr ]}=-{\text{e}}^{-z}f(z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af0c091271f23221b439e0bd828a493630db614e)
By the fundamental theorem of calculus, we get:

Now let:

Summing all the terms yields:
![{\displaystyle {\begin{aligned}\sum _{i\,=\,1}^{m}F(\beta _{i})-\sum _{i\,=\,1}^{m}{\text{e}}^{\beta _{i}}F(0)=\sum _{i\,=\,1}^{m}A_{i}\\[5pt]\sum _{i\,=\,1}^{m}F(\beta _{i})-F(0)\sum _{i\,=\,1}^{m}{\text{e}}^{\beta _{i}}=\sum _{i\,=\,1}^{m}A_{i}\\[5pt](2^{n}\!-m)F(0)+\sum _{i\,=\,1}^{m}F(\beta _{i})=\sum _{i\,=\,1}^{m}A_{i}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbc950f2267c807c94eb55c07eb8a139fe4d9e19)
Lemma: Let
be a polynomial with a root
of multiplicity
. Then
for all
.
Proof: By strong induction.
Let us write
, with
a polynomial such that
.
For
we get:

Assume that for all
the claim holds for all
.
We shall prove that for
the claim holds for all
:
![{\displaystyle {\begin{aligned}f(z)&=(z-z_{0})^{k+1}Q(z)\\[5pt]f^{(1)}\!(z)&=(k+1)(z-z_{0})^{k}Q(z)+(z-z_{0})^{k+1}Q^{(1)}\!(z)\\[5pt]&={\color {blue}(z-z_{0})^{k}}{\color {red}{\bigl [}(k+1)Q(z)+(z-z_{0})Q^{(1)}\!(z){\bigr ]}}\\[5pt]&={\color {blue}(z-z_{0})^{k}}{\color {red}R(z)}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc70fabb3477d5ddd7b0e5ac82177d364049d396)
The blue part is of multiplicity
, with
a polynomial such that
.
Hence their product satisfies the induction hypothesis.
Let us now define:
![{\displaystyle {\begin{aligned}f(z)&=(b_{m})^{c}g(z),\quad (c=mp-1)\\[5pt]g(z)&={\frac {1}{(p-1)!}}\,z^{p-1}{\bigl [}B(z){\bigr ]}^{p}\\[5pt]&={\frac {(b_{0})^{p}}{(p-1)!}}z^{p-1}+\sum _{k\,=\,p}^{p\,+\,c}{\frac {d_{k}}{(p-1)!}}z^{k},\quad (d_{k}\in \mathbb {Z} )\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/511c6913b5d0ad65f931b559da281a499be7991c)
and
is a prime number such that
. We get:

hence for all
, the function
is a polynomial with integer coefficients all divisible by
.
By parts 1 and 3, we get:
![{\displaystyle {\begin{aligned}N&=(2^{n}\!-m)F(0)+\sum _{i\,=\,1}^{m}F(\beta _{i})\\[2pt]&=(2^{n}\!-m)\sum _{j\,=\,0}^{p\,+\,c}f^{(j)}\!(0)+\sum _{i\,=\,1}^{m}\sum _{j\,=\,0}^{p\,+\,c}f^{(j)}\!(\beta _{i})\\[2pt]&=(2^{n}\!-m)\!\!\sum _{j\,=\,p-1}^{p\,+\,c}\!\!\!f^{(j)}\!(0)+\sum _{i\,=\,1}^{m}\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(\beta _{i})\\[2pt]&=(2^{n}\!-m)\,{\bigg [}\,f^{(p-1)}\!(0)+\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(0)\,{\bigg ]}+\sum _{j\,=\,p}^{p\,+\,c}\sum _{i\,=\,1}^{m}f^{(j)}\!(\beta _{i})\\[2pt]&={\color {red}(2^{n}\!-m)(b_{0})^{p}(b_{m})^{c}}\!+{\color {green}(2^{n}\!-m)\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(0)}+{\color {blue}(b_{m})^{c}\sum _{j\,=\,p}^{p\,+\,c}\sum _{i\,=\,1}^{m}g^{(j)}\!(\beta _{i})}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af8b641b7724f3f5584660edb1dd6a401cd40dbf)
The red part is an integer not divisible by
.
The green part is an integer divisible by
.
The blue part is the most important:
By Vieta's formulae we get

and the sums are symmetric polynomials in
. Therefore, these can be expressed as polynomials
![{\displaystyle {\begin{aligned}\sum _{i\,=\,1}^{m}g^{(j)}\!(\beta _{i})&=G_{j}({\vec {E}}{}^{m}({\vec {\beta }}{}^{m}))\in \mathbb {F} [{\vec {\beta }}{}^{m}]\\&={\frac {a_{j}}{(b_{m})^{c_{j}}}},\quad (a_{j},c_{j}\in \mathbb {Z} ,\,c_{j}\geq 0)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0663d81b293260ab853dce9ccd7fa2fc4787bec7)

In addition, we get:
![{\displaystyle {\begin{aligned}\deg(g^{(j)})\leq c&\implies \deg(G_{j})\leq c,\quad (p\leq j\leq p+c)\\[3pt]&\implies 0\leq c_{j}\leq c\\[3pt]&\implies (b_{m})^{c-c_{j}}\in \mathbb {Z} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84f96febdd832b75f694dcc3d2374528688b22bb)
Hence, the blue part is an integer divisible by
.
Conclusion:
is an integer not divisible by
, and particularly
.
By part 2, by the triangle inequality for integrals we get:
![{\displaystyle {\begin{aligned}|A_{i}|&={\Bigg |}\int \limits _{0}^{\beta _{i}}{\text{e}}^{\beta _{i}-w}f(w)dw\,{\Bigg |}\\[2pt]&\leq \int \limits _{0}^{\beta _{i}}\,{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}\,{\bigr |}f(w){\bigl |}\,|dw|\\[2pt]&=\int \limits _{0}^{\beta _{i}}{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}\left|{\frac {(b_{m})^{c}}{(p-1)!}}\,w^{p-1}{\bigl [}B(w){\bigr ]}^{p}\right||dw|\\[2pt]&=\int \limits _{0}^{\beta _{i}}{\frac {{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}}{|b_{m}|}}\,{\frac {|b_{m}|^{mp}|w|^{p-1}|B(w)|^{p}}{(p-1)!}}\,|dw|\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8fcb69ac91de7c0b2ade1e27e41242b2081097d)
By the triangle inequality, we get:

On the other hand, we get

hence for sufficiently large
we get
. A contradiction.
Conclusion:
is transcendental, hence
is transcendental.