Let $ P $ be a non-constant polynomial in z. Show that $ P(z) \rightarrow \infty $ as $ z \rightarrow \infty $ This is a a homework problem I have and I am having some trouble on it. I had thought I solved it, but I found out a algebraic mistake made my proof incorrect. Here is what I have so far.
Let $ P(z) = \sum\limits_{k=0}^n a_k z^k $
$\def\abs#1{\left|#1\right|}$
Then, $ \abs{a_nz^n} = \abs{\sum\limits_{k=0}^n a_k z^k - \sum\limits_{k=0}^{n-1} a_kz^k} \leq \abs{\sum\limits_{k=0}^n a_k z^k} + \abs{\sum\limits_{k=0}^{n-1} a_kz^k} $ by the triangle inequality.
Let $ M \gt 0, \exists R \in \mathbb{R} \text{ s.t. } \abs{z} \gt R \implies M \lt \abs{a_n z^n} $ since it is known that $ \abs{a_n z^n} $ converges to infinity. So by choose such and R as above and choosing a sufficiently large z, you get:
$ M \lt \abs{a_n z^n} \leq \abs{\sum\limits_{k=0}^n a_k z^k} + \abs{\sum\limits_{k=0}^{n-1} a_kz^k} \implies M - \abs{\sum\limits_{k=0}^{n-1} a_kz^k} \lt \abs{\sum\limits_{k=0}^n a_k z^k} $
However, this is not sufficient to prove the hypothesis and I am at a loss of what to do. Help woud be great. 

Hints:
$$(1)\;\;\;\;\;|P(z)|=\left|\;\sum_{k=0}^na_kz^k\;\right|=|z|^n\left|\;\sum_{k=0}^na_kz^{k-n}\;\right|$$
$$(2)\;\;\;\;\forall\,w\in\Bbb C\;\;\wedge\;\forall\;n\in\Bbb N\;,\;\;\left|\frac{w}{z^n}\right|\xrightarrow[z\to\infty]{}0$$
