I have to prove the following problem.
Let $a \in \mathbb{Z}$ prove if $a>1$ then for all $n \in \mathbb{N}$ with $n>1$, $a<a^{n}$
Base case since $a>0$, $(a)(a)>a \implies a^2>a$ thus true for $n=2$
Inductive step
Assume $a>1$ and $a< a^{k}$ for some $k \in \mathbb{N}$.
Since $a>1$ and $a>0$,
$a<(a)(a)<a(a^k)=a^{k+1} \implies a<a^{k+1}$ thus true for all $n \in \mathbb{N}$