Q:

If $1<a$ how would I prove $n(a^{n}-1)-1$ is positive for any positive integer n? Is it possible to prove with the information given?

I've deduced:
$a^{n}>1 \Rightarrow a^{n}-1>0 \Rightarrow n(a^{n}-1)>0 \Rightarrow n(a^{n}-1)>-1$.
Where do I go from here?

A:

I hope that the followig might be of interest:
Let $a>2$. Then there is $b>1$ such that $a=b+1$. Then we get, by the Bernoulli inequality:
$a^n=(b+1)^n \ge 1+nb > 1+n$. 
Hence $n(a^{n}-1) > n^2 \ge1$ and we derive
$n(a^{n}-1)-1>0.$

