Prove that $2^n>\frac{n(n-1)(n-2)}6$ for $n\in\Bbb N$ Apparently, I'm supposed to prove this using the binomial theorem, but that doesn't seem intuitive. Can't I just say

Prove that $\exists n\in\Bbb N$ such that
  $$2^n\le\frac{n(n-1)(n-2)}6$$

And show that $n=1$ contradicts this? Is that a valid proof?

$$
2^n = (1+1)^n = 1 + n + \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} + \cdots + \frac{n(n-1)(n-2)\ldots 3.2.1}{n!}
$$
$$
= \frac{n(n-1)(n-2)}{3!} + \text{other positive terms} > \frac{n(n-1)(n-2)}{6}
$$
