How to prove $a^n < n!$ for all $n$ sufficiently large, and $n! \leq n^n$ for all $n$, by induction?

HINT $\ $ Rewrite them as products so to make the induction completely trivial.

Namely $\quad\quad\displaystyle \frac{n^n}{n\:!}\ =\ \ \: \frac{n}1\ \ \ \frac{n}2\: \ \ \frac{n}3\ \:\cdots\ \frac{n}n\ \ge\ 1\ \ $ since each factor is $\:\ge 1$

$\quad$ and $\rm\quad\quad\displaystyle\ \frac{n\:!}{5^n}\ =\ \frac{12\:!}{5^{12}}\ \frac{13}{5}\ \frac{14}5\ \cdots\ \frac{n}5\ >\ 1\ \ $ since each factor is $\: > 1$

Now you need only inductively prove the lemma that a product is $\ge 1\:$ or $> 1\:$ if each factor is - which is completely trivial. Many inductive proofs can be similarly drastically simplified by conceptual preprocessing. Here we've effectively employed multiplicative telescopy to reduce generally intractable exponential inequalities to tractable polynomial inequalities. $\: $ This is a powerful technique with widespread applications. See also the analogous case of additive telescopy and the fundamental theorem of difference calculus.

Your first proof is correct. (If $a\le c$ and $b\le d$ and they're all positive, then $ab\le cd$.) Your second one needs just a tiny bit more. How do we know that $n!\le (n+1)^n$? It follows from your inductive hypothesis, but it's not completely trivial.

As a general tip, try not to work your proofs backwards. When you're trying to prove that $P(k)\Rightarrow P(k+1)$, you start out by stating $P(k+1)$ and then deriving a true statement. It works here because pretty much everything you're doing is reversible (dividing by $k+1$ vs. multiplying), but in later life, this might no longer be true and so it's a good practice to start now.

Otherwise, good work!