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?
On
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 $m$.
$\blacksquare$
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 $n$.