Proof that for any $a>1$, $a^n > 1 +na$ for sufficiently large $n$

Question

I am looking for an elementary proof (preferably using nothing more advanced than the ordered field properties of $\mathbb R$) that for any $a>1$, $a^n > 1+na$ for a sufficiently large natural number $n$. Obviously one can prove this using the Taylor series for $a^x$, or by l'Hospital's rule, or other techniques, but I'm hoping for something that uses nothing more than the basic idea of $a^n$ as repeated multiplication. I feel that there must be a simple induction proof but it is eluding me at the moment.

Answer 1

By Binomial Theorem $a^{n} =((a-1)+1)^{n} > 1+\frac {n(n-1)} 2 (a-1)^{2}$. So it is enough to show that $\frac {n(n-1)} 2 (a-1)^{2} >na$ for $n$ sufficiently large. Take $n >1+\frac {2a} {(a-1)^{2}}$