Proof
Suppose for the sake of contradiction that $(x^n)_{n=1}^{\infty}$ converges to some limit $L$. Consider the identity $(1/x)^n (x^n) = 1$. Since this holds for all $n \in \mathbb{N}$, $\lim_{n \rightarrow \infty}(1/x)^n \lim_{n \rightarrow \infty} x^n = 1$, and $\lim_{n \rightarrow \infty} (1/x)^n = 1/L$. But $\lim_{n \rightarrow \infty} (1/x)^n = 0$ if $x > 1$, a contradiction. As such, $\lim_{n \rightarrow \infty}(x^n)_{n=1}^{\infty}$ diverges when $x>1$.
A more reasonable argument would be to use the inequaiity $x^{n}=(1+(x-1))^{n} >1+n(x-1) \to \infty$.