Prove for any $a>1$ there exist a natural number, $N$, such that $a^n>n$ for all $n \ge N$
By definition, for $N$ to be a natural number $1\le N \le ∞$ for any integer $N$.
By induction, the base case $n=1$ is true because $a^1>1$ which is true.
Now assume that $a^k>k$ is true, then it is also true for $n= k+1$.
Now I am not sure if I am on the right track