$2^n > n^4$ 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?

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 $n\ge 16$.
