Prove that $2^n>n^4$ for all $n\geq 17$ I'm always a bit fuzzy on how to solve induction problems involving inequalities. I've managed to get somewhere though, but it looks like I have to go down four levels of induction to prove.
This is what I have so far:
Base ($n=17$):
$$2^{17}>17^4\Rightarrow131072>83521$$
Step:
Assumption: $2^k>k^4$
\begin{align}
  2^{k+1}&>(k+1)^4\\
  2\cdot 2^{k}&>(k+1)^4\\
  2^k+2^k&>k^4+4k^3+6k^2+4k+1
\end{align}
From here, it looks like I need to claim that $2^k+2^k>k^4+2^k$, but in order to do this, I would then need to prove that $2^k>4k^3+6k^2+4k+1$, and then again probably one or two more times.
Is there a different approach I can take other than going down this long and tedious route?

all that is required for the induction step is $$ (n+1)^4 < 2 n^4 $$
for large enough $n,$ so
$$ \frac{(n+1)^4}{  2 \, n^4} < 1. $$
This is the same as 
$$ \left( \frac{n+1}{n} \right)^4 =  \left( 1 + \frac{1}{n} \right)^4 < 2 $$
This does not work for $n \leq 5,$ indeed
 $(6/5)^4 = 1296/ 625,$ but it does work for $n=6$ as $(7/6)^4 = 2401/1296$ and $2 \cdot 1296 = 2592.$
Meanwhile  $$  =  \left( 1 + \frac{1}{n} \right)^4  $$
is strictly decreasing in $n,$ so it is less than $2$ for all $n \geq 6.$ It may help to consider $n=10,$ as $(11/10)^4 = 14641/10000 = 1.4641 < 2.$
If $n \geq 17$ and $$ \frac{n^4}{2^n} < 1, $$ then also $n \geq 6,$ so when $n \geq 17,$
$$ \color{red}{\frac{(n+1)^4}{2^{n+1}} = \frac{n^4}{2^n} \cdot \frac{(n+1)^4}{2 n^4} <  \frac{n^4}{2^n} < 1} $$
No real tricks here: if you have positive things $A,B$ that depend on some $n,$ and you want to show $A < B,$ you can try for $A-B < 0$ or for $\frac{A}{B} < 1.$ Either way might give you something simpler than separately considering $A,B.$ The induction step then becomes showing that the left hand side decreases as $n$ is increased to $n+1.$
