Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$ For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one: 

Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$

My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$.
Note that $2^n=(1+1)^n$. Thus by the Binomial Theorem we have that, 
$
(1+1)^n=\sum_{k=0}^{n} {n \choose k}(1)^{n-k}(1)^n
$
Evaluating some of the first couple I get the following terms:
$1+n+\dfrac{n(n-1)}{2}+\dfrac{n(n-1)(n-2)}{6}+\dfrac{n(n-1)(n-2)(n-3)}{24}+...$
I have noticed that for $n<4$, $n^3 \geq 2^n$. However, it seems that when $n\geq 4, n^3 < 2^n$. Thus, would it be possible to make an argument that for $n \geq 4, s_{4}>s_{5}>s_{6}>...$? Therefore, by evaluating for $n=4$ and using some algebra you get the following:
$s_{4}=\dfrac{24n^3}{24+24n+4n(n-1)(n-2)+n(n-1)(n-2)(n-3)} \leq \dfrac{24n^3}{n^4} = \dfrac{24}{n}$ 
and we know that lim$_{n \rightarrow \infty} \left( \dfrac{1}{n} \right)=0$ and lim$_{n \rightarrow \infty} \left( -\dfrac{1}{n} \right)=0$
Therefore, since $\dfrac{1}{n} \geq s_{4} > s_{5} >s_{6} > ...s_{n}\geq-\dfrac{1}{n}$, by the Squeeze Theorem $s_{n}$ converges to zero as well?

Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.
