Prove that $4^n > n^4$by induction for $n\ge 5$ That is a simple question and I can't start a simple desenveloment.
Just $k=5$ we have $4^5 = 1024 > 5^4 = 625$
for $k+1$: $4^{k+1} > (k+1)^4\Rightarrow 4^k > (k+1)^4/4$
And How can i proceed after that?
Best,
 A: A different approach using calculus (not induction)
Let $f: \mathbb R_+^* \supset \mathbb N^* \to \mathbb R$ such that $f(x) = \ln x / x$ which has negative derivative for every $x>e$. So
$$n \geq 5 > 4 > e \iff f(n) < f(4) \iff \ln n / n < \ln 4 / 4 \iff \ln (4^n) > \ln (n^4) \iff 4^n > n^4$$ which extends as well to the natural numbers
A: Start with the induction assumption, 
$$n^4 <4^n$$ multiply both sides by $4$ to get 
$$4n^4 < 4^{n+1}$$
now show that 
$$(n+1)^4 \leq 4 n^4$$ in which $n\geq 5$ is used.
An easy way to show this last is to prove 
$$\left(1+\frac{1}{n}\right)^4 < 4$$ since 
$$1+\frac{1}{n} \leq 1+\frac{1}{5}$$ we just have to check, 
$$\left(\frac{6}{5}\right)^4 < 4$$
and this is true.
A: Hint:
For $k \geq 5$ the following is true:
$$k\sqrt2 > k+1$$
As:
$$5(\sqrt2 -1)>5\times \frac25 =2> 1$$
We can show that $\sqrt2 -1 >\frac25$ as such:
$$\sqrt2>\frac75$$
Now as both $\sqrt2$ and $\frac75$ are bigger than zero, the inequality is persevered by squaring.
$$2=\frac{50}{25}>\frac{49}{25}$$
