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?

The proof you suggest would not work- you've shown that for $n=1$, that inequality holds, but that doesn't prove anything about any other natural number. This is a good time for one of my favorite types of proofs- the combinatorial proof. As you mentioned, the binomial theorem can be used here, as $\frac{n(n-1)(n-2)}{6}=\frac{n!}{(n-3)!3!}={n\choose n-3}$. Think of the combinatorial interpretation of $n\choose n-3$, especially with respect to sets. It counts a specific type of subsets of a set of size $n$. You also need to think of $2^n$ with respect to sets, as it also counts subsets of a set of size $n$.
