Prove by induction the inequality $2^{n!}>n^n$ for $n>2$ My attempt so far:
Base case:
$2^{3!}>3^{3}$
$2^6 > 27$
$64> 27$
Then going for
$2^{(n+1)!} > (n+1)^{(n+1)}$
I get
$(2^{(n!)})^{(n+1)} > (n+1)^{(n+1)}$
Since n+1 is positive,
$2^{n!} > n+1$
And I do not go where to go from here. I am unsure if I am following the correct path. Am I approaching this in the correct way?
 A: We're trying to prove $2^{n!} > n^n$. You've shown that the base case holds - great.
In an induction proof you assume that the statement holds for an arbitrary $k$, and then you have to show it holds for $k+1$ dependent on it holding for $k$.
$2^{(k+1)!} = 2^{(k+1)k!} = (2^{k!})^{k+1}$
Now we have to remember that we're assuming that the statement holds for $k$, ie we assume $2^{k!} > k^k$. Also we clearly have $k^k > k+1$. Putting these two inequalities together and raising to the $k+1$'th power we get
$$(2^{k!})^{k+1} > (k^k)^{k+1} > (k+1)^{k+1}$$
and so
$$2^{(k+1)!} > (k+1)^{k+1}$$
which concludes the proof.
Essentially we prove that if it's true for a certain $k$, then it will also be true of $k+1$ and since you yourself showed that it's true for the lowest $n$ in the set you're considering, it must be the case that it's true for any integer higher then that $n$.
A: Essentially the solution above is right. I'll give the same but with an extra.
BC (Base case)
You've already proven this.
IH (Induction Hypothesis)
Suppose for some integer $m \ge 3$ that
$$ 2^{m!} > m^{m}$$
You can notice now, that
$$ 2^{m!} > m^{m} > m+m > m+1 $$
And thus, $2^{m!} > m+1 $. Finally, if we raise both sides to the power $ m+1$ , we get.
$$ 2^{(m+1)!} > (m+1)^{m+1} $$
Which of course will hold for every integer $m$.
$\blacksquare$
