Is there a whole number $x\in\mathbb{Z}$ with $x\neq 0$ s.t. $\exp(x)$ is natural?

Question

I was wondering if there is a number $x\in\mathbb Z$ with $x\neq 0$ s.t. $\exp(x)\in\mathbb N$ and if not, why is that so?

EDIT: Forgot to exclude the $0$.

Answer 1

Besides $0$, no. Suppose that $\exp(n)\in \mathbb{Z}$, where $n\neq 0$. Then $e=\exp(n)^{1/n}$ is algebraic over $\mathbb{Q}$, contradicting the fact that $e$ is transcendental.

Note that this can be expanded to show that, moreover, $\exp(r)$ is not rational for any non-zero rational number $r$, by following the same argument.

Answer 2

$$\exp(0)=1$$ $\qquad$$\qquad$$\qquad$$\qquad$

Answer 3

Suppose $e^x=y$, $x\in \mathbb Z-\{0\}$, $y\in \mathbb N$, then $e$ is the solution of $z^x-y=0$, a polynomial equation with coefficient in $\mathbb Z$, contradict with the fact that $e$ is a transcendental number.