Proving that $3^n \gt n^4 \ \forall n \gt 8$ Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:

Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$
Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$
Inductive step ($P(n) \implies P(n+1)$): $\  3^{n+1} \gt  \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$
$$3^{n+1} = 3^n + 3 = 3^n + 3^n + 3^n = 3^n + 3^n + \frac{(n+1)^4}{3} \gt (n+1)^4 \\ \text{I found } \frac{(n+1)^4}{3} \text{ by dividing 3 on both sides of } 3^{n+1} \gt  (n+1)^4 \text{ as soon as I suppose this to be true.}$$

 A: Rewrite the expression as $e^{n \log 3 - 4 \log n}$. This is your induction step. Now you need to show it for $n+1$.
$$
e^{(n+1)\log 3 - 4 \log (n+1)} = e^{n \log 3 - 4 \log n} \cdot e^{\log 3 - \log (1+\frac{1}{n})}
$$
The first term is greater that $1$ by the inductive argument. All is left to show that the second term is greater than $1$. This is true (again, using the definition of logarithm) for
$$
\frac{3n}{n+1} >e
$$
which is true for $n>\frac{e}{3-e} \approx 1.5$
A: 
Inductive step ($P(n) \implies P(n+1):) \  3^{n+1} \gt  \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$

You need to PROVE $ 3^{n+1} \gt  \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$.  You can NOT start by assumenting that.
But you start by assuming $3^n > n^3$.
So $3^{n+1} = 3\cdot 3^n > 3n^4$
So how does $3n^4$ compare with $(n^4 + 4n^3 + 6n^2 + 4n + 1)$?
Well $3n^4 = n^4 + 2n^4$.
SO how does $2n^4$ compare with $4n^3 + 6n^2 + 4n + 1$?
Well $n \ge 8$ so $2n^4 = (2n)n^3 \ge 8n^3$.
So
$n^4 + 2n^4 \ge n^4 + 8n^3 = n^4 + 4n^3 + 4n^3$.
So how does $4n^3$ compare with $6n^2 + 4n + 1$.
Well, we just keep doing this $n \ge 8$ so $n^k \ge 8n^{k-1}$ to get:
$3^{n+1} = 3\cdot 3^n >$
$3n^4 = n^4 + 2n^4 \ge $
$n^4 + 8n^3 = n^4 + 4n^3 + 4n^3 \ge$
$n^4 + 4n^3 + 32n^2= n^4 + 4n^3 + 6n^2 + 26n^2\ge$
$n^4 + 4n^3 + 6n^2 + 26\cdot 8 n=$
$n^4 +4n^3 + 6n^2 + 4n + (26\cdot 8 - 4)n \ge$
$n^4 + 4n^3 + 6n + 4n + (26\cdot 8 -4)\cdot 8 >$
$n^4 + 4n^3 + 6n + 4n + 1=(n+1)^4$
