TITLE: Prove that $n^k < 2^n$ for all large enough $n$
QUESTION [7 upvotes]: If $k\ge 2$ is an integer,
prove by elementary means
(no calculus or limits)
that
there is a $N(k)$
such that
$n^k < 2^n$
for all integers $n \ge N(k)$.
Give an explicit form for $N(k)$.

REPLY [2 votes]: Pick $n > 2k$. Then, 
$$
2^n > \binom{n}{k + 1} = \frac{n (n - 1) \cdot \ldots \cdot (n - k)}{(k+1)!}
\geq \frac{n^{k + 1}}{2^{k+1} (k+1)!}
$$ 
by a trivial combinatorial argument for the first inequality, and using $n - k > n/2$ in the second inequality.
Therefore $2^n > n^k$ as soon as
$$
n > 2^{k + 1} (k + 1)!
$$