I have been experimenting with the lambert W function ($W(x)$) and transcendental numbers, and have attempted to prove this result, however I am unable to find any confirmation that $e^{W(\pi)}$ is transcendental and so am doubting my proof. It goes as follows:
By definition of Lambert W function
$W(\pi)e^{W(\pi)}=\pi$
$e^{W(\pi)}=\frac{\pi}{W(\pi)}$
${(e^{\pi})}^{\frac{W(\pi)}{\pi}}=\frac{\pi}{W(\pi)}$
As $e^{i{\pi}}=-1$
$e^\pi=(-1)^{-i}$
$(-1)^{i\frac{W(\pi)}{\pi}}=\frac{W(\pi)}{\pi}$
We have three cases:
1) $\frac{W(\pi)}{\pi}$ is rational
2) $\frac{W(\pi)}{\pi}$ is irrational algebraic
3) $\frac{W(\pi)}{\pi}$ is irrational transcendental
Also $a^b$ is transcendental where a is algebraic but not 0 or 1, and b is irrational algebraic (by the Gelfond–Schneider theorem)
In case 1, if $\frac{W(\pi)}{\pi}$ is rational, $i\frac{W(\pi)}{\pi}$ is irrational algebraic, so by the Gelfond–Schneider theorem $\frac{W(\pi)}{\pi}$ is transcendental, so there is a contradiction
In case 2, if $\frac{W(\pi)}{\pi}$ is irrational algebraic, $i\frac{W(\pi)}{\pi}$ is also irrational algebraic as $\frac{W(\pi)}{\pi}$ is real and i is imaginary, so by same logic there is a contradiction
Therefore $\frac{W(\pi)}{\pi}$ must be transcendental
Finally as $W(\pi)e^{W(\pi)}=\pi$
$\frac{W(\pi)}{\pi}e^{W(\pi)}=1$
And as $\frac{W(\pi)}{\pi}$ is transcendental, $e^{W(\pi)}$ is also transcendental