Was just hoping to validate my proof: Let $n=16\Rightarrow 2^{16}>16^4\Rightarrow 65536\not\gt65536$. Prove $P(n)=2^n>n^4$ whenever $n>16$.
Basis Step: $P(17)=2^{17}>17^4\Rightarrow 131072>83521\equiv T$
Inductive Step: $P(k)=2^k>k^4$, assume $P(k)$ is true for all $k>16$
Prove: $P(k+1)=2^{k+1}>(k+1)^4$ $$(k+1)^4=k^4 + 4 k^3 + 6 k^2 + 4 k + 1\le k^4 + 4 k^3 + 6 k^2 + 16k $$ $$\le k^4 + 4 k^3 + 16k^2 $$ $$\le k^4 + 16 k^3 $$ $$\le k^4+k^4= 2k^4$$ $$2\cdot 2^k>2k^4\Rightarrow 2^{k+1}>(k+1)^4$$
Note that $P(k)$ is the assertion “$2^k>k^4$”. So, it is not a good a idea to write$$P(k)=2^k>k^4.$$It looks as if you are asserting that $P(k)=2^k$.
The idea of your proof is correct, although it is a bit hard to read. It would be simpler to note that$$\frac{2^{k+1}}{2^k}=2$$and that, if $k>16$,$$\frac{(k+1)^4}{k^4}=\left(1+\frac1k\right)^4<\left(1+\frac1{16}\right)^4<2.$$