TITLE: Help getting started on a proof
QUESTION [5 upvotes]: I honestly have no idea where to start with the following proof, and I was wondering if anyone could help me get started. I don't want the whole idea, I just need to know where to start with this proof. 
The question:
Let x be a real number. Suppose 1 < x. Given any M > 0; prove there is a natural number N such that n $\geq$ N implies M $\leq$ $x^n$.
My book has absolutely no examples (it's a custom text), and I honestly have just stared at this problem for a few hours now. If someone would be willing to just help me get going, I think I should be able to work out the rest from there. I appreciate any help anyone would be willing to offer, thank you.
edit: Here is the solution I came up with after the generous help I received below.
Let M > 0 and let x = 1 + t. Since x > 1, we have 1 + t > 1 if and only if t > 0. We then have $x^n$ = $(1 + t)^n$. Using the Bernoulli Inequality, we have $(1 + t)^n$ $\geq$ 1 + nt for all n $\geq$ 0. We then have $x^n$ $\geq$ 1 + nt > nt, thus $x^n$ > nt for all n $\geq$ 1. Let N be a natural number with N = $\frac{M}{t}$. Then n $\geq$ N = $\frac{M}{t}$ implies nt $\geq$ M. Since $x^n$ > nt, we have that n $\geq$ N implies $x^n$ > M.
I know my proof-writing skills are very bad and I'm sure there are a million ways I can improve what I've done, but I am very thankful for the help I received here from Andre Nicolas.

REPLY [4 votes]: A start: Let $x=1+t$. Then $t\gt 0$ and $x^n=(1+t)^n$. Now use the Bernoulli Inequality 
$$(1+t)^n \ge 1+nt.$$ This can be proved by induction, or by using the Binomial Theorem.