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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 $n \in \mathbb{N}$

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    $\begingroup$ Your argument is correct :) $\endgroup$ – Reveillark Dec 2 '19 at 5:08
  • $\begingroup$ I might be reading your proof incorrectly, but at the base case you state that $a>0$ implies that $(a)(a) > a$. This is only true when $a>1$. $\endgroup$ – Axion004 Dec 2 '19 at 5:46

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