Prove that $3^n>n^4$ if $n\geq8$ Proving that $3^n>n^4$ if $n\geq8$
I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove $n=k+1$.  Any ideas?
Thanks!
 A: HINT: Show that if $k\ge 8$, then
$$3\ge\left(\frac{k+1}k\right)^4=\left(1+\frac1k\right)^4\;.$$
When you go from $k$ to $k+1$, you’re multiplying $3^k$ by $3$, and you’re multiplying $k^4$ by $\left(\frac{k+1}k\right)^4$.
A: Hint:
Look at $(n+1)^4=n^4+4n^3+6n^2+4n+1$.
Notice that $4n^3 +6n^2\leq (n/2)n^3+n^3\leq n^4/2+n^4/2=n^4$
and $4n+1\leq n^4$.
Thus you get $(n+1)^4\leq n^4+n^4+n^4$.
Now what? (You should be able to take it from here.)
A: You want to show $3^n>n^4$. This i.e. to showing $e^{n\ln3}>e^{4\ln n}$. This means you want to show  $n\ln 3>4\ln n$. It suffices to show $\frac{n}{\ln n }>\frac{4}{\ln 3}$. Since $\frac{8}{\ln 8}>\frac{4}{\ln 3}$ and since $f(x)=\frac{x}{\ln x}$ has a positive first derivative for $x\geq 8$, the result follows. 
A: First show the base case. Then assume $3^k>k^4$. Hence, $3^{k+1}=3*3^k>3k^4>(k+1)^4$
A: It can be proved by induction. The n+1 can be solved in the following way (assuming the claim is correct for n): need to be prove $3^{n+1}\geq (n+1)^4$. but we assumed that $3^n\geq n^4$ so the claim also can be written as $3*3^n\geq3*n^4\geq(n+1)^4$ but taking the 4th square on each side (both sides are non-negative), $3^\frac{1}{4}n\geq n+1\rightarrow 0.31n\geq 1$ which is correct since we prove the claim for $n\geq 8$.
