Show that $e^{\sqrt 2}$ is irrational

I'm trying to prove that $e^{\sqrt 2}$ is irrational. My approach: $$ e^{\sqrt 2}+e^{-\sqrt 2}=2\sum_{k=0}^{\infty}\frac{2^k}{(2k)!}=:2s $$ Define $s_n:=\sum_{k=0}^{n}\frac{2^k}{(2k)!}$, then: $$ s-s_n=\sum_{k=n+1}^{\infty}\frac{2^k}{(2k)!}=\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{\prod_{k=1}^{2k}(2n+2+k)}\\<\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{(2n+3)^{2k}}=\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2} $$ Now assume $s=\frac{p}{q}$ for $p,q\in\mathbb{N}$. This implies: $$ 0<\frac{p}{q}-s_n<\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2}\iff\\ 0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2} $$ But $\left(p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}\right)\in\mathbb{N}$ which is a contradiction for large $n$. Thus $s$ is irrational. Can we somehow use this to prove $e^\sqrt{2}$ is irrational?

Solution 1:

Since the sum of two rational numbers is rational, one or both of $e^{\sqrt{2}}$ and $e^{-\sqrt{2}}$ is irrational. But, $e^{-\sqrt{2}}=1/e^{\sqrt{2}}$, and hence both are irrational.

Solution 2:

$e^{\sqrt{2}}$ is transcendental because of Lindemann–Weierstrass theorem:

If $a\neq 0$ is algebraic, then $e^a$ is transcendental.

It is written in the list of transcendental numbers.

Solution 3:

To prove that $e^{\pm\sqrt{n}}\not\in\mathbb{Q}$, it is enough to exploit the Gauss' continued fraction for $\tanh$ and Lagrange's theorem about the periodicity of continued fractions of quadratic irrationals.
A detailed explanation of this approach (for $n=2$) is outlined in this answer.