Can We prove Transcendence of $e$ if we know that $e^{m}$ is Irrational?

Question

So I just managed to prove that $e^{m}$ where $m$ is an integer is irrational. I was wondering If I can prove using this fact that $e$ is Transcendental. The only thing I came up with was that if we set this polynomial $P(x)=c_{0}+c_{1}x+c_{2}x^2+\cdots+c_{n}x^{n}$ where $c$ is an integer and then assume that $P(e)=0$ we're saying that: $c_{1}e+c_{2}e^{2}+c_{3}x^{3}+\cdots+c_{n}e^n=-c_0$. Of course every term on left is an Irrational Number. On the right side we have an Integer. If we could only proof that the left side is an Irrational Number We would have a needed contradiction and thus prove that e is an Transcendental Number.

Answer 1

Fix $x=\sqrt2+1$.

Then $x^m$ is irrational for every $m>0$, but $x$ is not transcendental.

Answer 2

Note that $y=\sqrt 2+\sqrt 3$ has the property that $y^m$ is irrational for every non-zero rational integer. And $y$ is algebraic, not transcendental. So your condition is not strong enough to prove your intended conclusion.