Prove that $e^n$ is irrational for any natural number $n$

Question

I have a question and it will be appreciated that you tell me some more details. Here is the question. For an arbitrary natural integer n, prove for any $n$, $e^n$ is irrational.

Answer 1

Let $n$ be a natural number. If $e^n$ were rational, say $e^n=\frac{p}{q}$, then $e=\sqrt[n]{\frac{p}{q}}$ would be an algebraic number (i.e., it is a zero of the polynomial $p(x)=qx^n-p$ whose coefficients are integers). But it is known that $e$ is not algebraic but transcentendal.