[conistonwater]

As pointed out in the annotation, isn't the case $e^2$ just exactly the same proof but for the special case $c=0$? And in any case, if $e^2$ were rational, then $e$ could only be rational or a quadratic irrationality.

[siegelzero]

Yeah, this proof seems to work. I was thinking of the case where you attack $e^2$ directly, instead of going after both $e$ and $e^{-1}$.

[pmalynin]

It can be shown by clever application of Taylor Expansion that e^2 is irrational.