$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
 A: As others have noted, your induction proof ultimately suffers from establishing what your base case is. Your deduction that
$$
2^{1/4}(n)>n+1
$$
requires $n\geq6$, as Barry notes, but your inequality is not even true until $n\geq17$, as lulu notes. This means that your base case should be, at minimum, for when $n=17$. Then you can proceed exactly as you have done.

"Is there a better way to solve this problem?" Sure: use the fact that any exponential function eventually overtakes any power function (with bases $>1$ of course). But that is not all that insightful here--perhaps you are after something more intuitive perhaps? If so, please specify. 
A: There is another inductive way, I don't know if better or worse:
Our goal is to prove that $2^{n+1}>(n+1)^4$, assuming that $2^n>n^4$ and, as noted, $n\ge 17$. Let's estimate $2^{n+1}-(n+1)^4$:
$$2^{n+1}-(n+1)^4=\big(2^n-[(n+1)^4-n^4]\big)+(2^n-n^4)>(2^n-n^4)+(2^n-n^4)>0$$
To show the first inequality we have to check:
$$(n+1)^4-n^4<n^4$$
but
$$\left(\frac{n+1}n\right)^4\le\left(\frac{18}{17}\right)^4<2$$
A: Using the function  $x^{1/4}$ is not a purely algebraic proof. Here is one. Explicitely, we'll prove $2^n>n^4$ for all $n>16$.
For that, we'll prove by induction that if $n\ge 16$ and $2^n\ge n^4$, then $2^{n+1}>(n+1)^4$.
For $n=16$, we have an equality: $2^{16}=16^4$.
Now suppose that, for some $n\ge 16$, we have $2^n>n^4$. Then $2^{n+1}=2\cdot 2^n\ge 2n^4$. So we have to prove $2n^4>(n+1)^4$, i.e.
$$n^4>4n^3+6n^2+4n+1.$$
As $n>1$, $\;4n^3+6n^2+4n+1>4n^3+6n^3+4n^3+n^3=15n^3$. 
So it is enough to prove $n^4>15n^3$, which is equivalent to $n>15$. We've supposed $n\ge 16$.
A: Here is a nice little non-induction proof that $n^4\lt2^n$ for $n\gt16$.
Note that
$$n^4=\left(\sqrt n\over2\right)^8\cdot2^8\cdot1^{n-16}$$
So if $n\gt16$, the arithmetic-geometric mean inequality tells us
$$\sqrt[n]{n^4}\lt{8\cdot\left(\sqrt n\over2\right)+8\cdot2+(n-16)\cdot1\over n}={4\over\sqrt n}+1\lt2$$
hence $n^4\lt2^n$.  
(Remark: for $n=16$ both inequalities in the displayed expression become equalities.  In particular, all $16$ terms in the AGM product $\left(\sqrt4\over2\right)^8\cdot2^8$ are equal to $2$.  For $n\lt16$, the AGM connection falls apart, as it must, since the inequality goes the other way.)
