<s>[INST] Prove $3^n$ > $n^4$ if $n\geq 8$ As the title states, I am tasked with proving $3^n$ > $n^4$ if $n\geq 8$
The base case is trivial to prove. It is obvious that $3^8 > 8^4$ since
$3^8 = 6561$ and $8^4 = 4096$, and $6561 > 4096$, thus the base case $n = 8$ is true.
Now for the Induction Hypothesis (IH) we let $n = m$ which then says $3^{m} > m^{4}$
I know that $3^{m+1} > (m+1)^{4}$ is what we are trying to show. So when you expand this you get $$3*3^{m} > m^{4} + 4m^{3} + 6m^{2} + 4m + 1$$ (Binomial Expansion Theorem is how you get the right hand side of the above inequality). 
But this is where I'm stuck. I know that $3^{m} > m^{4}$ via the IH but idk what else to say or where to go. Any help would be appreciated. Thanks!
 [/INST] Pulling it together into a compact induction proof:
Base case: $3^8 = 9^4>8^4$
Inductive hypothesis: $3^k >k^4$ for some $k\ge 8$
$\begin{align}
\text{Observe }(k+1)^4 &= k^4 + 4k^3 + 6k^2 + 4k + 1 \\
&< k^4 + (4+6+4+1)k^3 \\
&< k^4 + 16k^3 \\
& \le 3k^4
\end{align}$
Then $3^{k+1}=3\cdot3^k>3\cdot k^4> (k+1)^4$ as required
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