$2^n > n^4$ 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?

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
$$\left(\frac{n+1}n\right)^4\le\left(\frac{18}{17}\right)^4<2$$
