Just for fun: it is enough to exploit the prime number theorem $\pi(n)\sim\frac{n}{\log n}$. Legendre's theorem on $\nu_p(n!)$ ensures
$$ n! = \prod_{p\leq n} p^{\lfloor\frac{n}{p}\rfloor+\lfloor\frac{n}{p^2}\rfloor+\ldots}\leq \prod_{p\leq n}p^{\frac{n}{p-1}} $$
which is a rather crude bound, but already sufficient:
$$ \log(n!) \leq n\sum_{p\leq n}\frac{\log(p)}{p-1}=n\sum_{m\leq n}\mathbb{1}_p(m)\frac{\log m}{m-1}\stackrel{\text{SBP}}{=}\underbrace{n\pi(n)\frac{\log n}{n-1}}_{O(n)}+n\sum_{m\leq n-1}\pi(m)\left[\frac{\log m}{m-1}-\frac{\log(m+1)}{m}\right].$$
Here $\mathbb{1}_p$ is the characteristic function of primes, $\pi$ is the prime-counting function and $\text{SBP}$ stands for summation by parts. In the last sum the main term behaves like $\frac{1}{m}+O\left(\frac{\log m}{m^2}\right)$, hence
$$\log(n!) = O(n)+n\sum_{m\leq n-1}\frac{1}{m} =n\log(n)+O(n).$$
Of course this is extreme cheating: historically speaking, weak forms of the PNT were derived from the asymptotics for $n!$ or $\binom{2n}{n}$, while here we performed just the opposite. A more elementary approach is to consider that
$$ \frac{4^n}{\sqrt{n+1}}<\binom{2n}{n}=\frac{(2n)!}{n!^2}<4^n $$
where the inequality on the right is trivial and the inequality on the left follows from $\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^2$ and Cauchy-Schwarz. By the theory of moments, the previous inequality holds for non-integer values of $n$, too, so by letting $L(n)=\log(n!)$ we have
$$ L(n)-2 L(n/2) < n\log(2), $$
$$ 2L(n/2)- 4L(n/4) < n\log(2), $$
$$ 4L(n/4)- 8 L(n/8) < n\log(2), $$
$$ \ldots $$
and
$$ L(n) - 2^k L\left(\frac{n}{2^k}\right) \leq kn\log(2). $$
By picking $k\approx \log_2(n)=\frac{\log(n)}{\log(2)}$ the claim $L(n)\sim n\log n$ is proved.
