[proofplan]
We express $\log(n!)$ as the sum $\sum_{k=1}^n \log k$ and bound it above and below by comparing the sum with the integral $\int_1^n \log x \, d\mathcal{L}^1(x)$. Both bounds are asymptotic to $n \log n$, so the squeeze gives $\log(n!) \sim n \log n$.
[/proofplan]
[step:Write $\log(n!)$ as a sum and bound it below by an integral]
Since $\log$ is an increasing function,
\begin{align*}
\log(n!) = \sum_{k=1}^n \log k \ge \int_1^n \log x \, d\mathcal{L}^1(x),
\end{align*}
where the inequality holds because $\log k \ge \log x$ for all $x \in [k-1, k]$, so $\sum_{k=2}^n \log k \ge \sum_{k=2}^n \int_{k-1}^k \log x \, d\mathcal{L}^1(x) = \int_1^n \log x \, d\mathcal{L}^1(x)$, and adding $\log 1 = 0$ on the left does not change the sum.
[guided]
The sum $\sum_{k=1}^n \log k$ is a left-endpoint Riemann sum for the increasing function $\log$. For an increasing function $f$, the left-endpoint Riemann sum over unit intervals satisfies $f(k) \ge f(x)$ for $x \in [k-1, k]$. Integrating both sides over $[k-1,k]$ gives $f(k) \cdot 1 \ge \int_{k-1}^k f(x) \, d\mathcal{L}^1(x)$. Summing from $k = 2$ to $n$:
\begin{align*}
\sum_{k=2}^n \log k \ge \sum_{k=2}^n \int_{k-1}^k \log x \, d\mathcal{L}^1(x) = \int_1^n \log x \, d\mathcal{L}^1(x).
\end{align*}
Since $\log 1 = 0$, the left side equals $\sum_{k=1}^n \log k = \log(n!)$.
[/guided]
[/step]
[step:Bound $\log(n!)$ above by an integral]
Similarly, since $\log$ is increasing, $\log k \le \log x$ for $x \in [k, k+1]$. Summing from $k = 1$ to $n$:
\begin{align*}
\log(n!) = \sum_{k=1}^n \log k \le \sum_{k=1}^n \int_k^{k+1} \log x \, d\mathcal{L}^1(x) = \int_1^{n+1} \log x \, d\mathcal{L}^1(x).
\end{align*}
[/step]
[step:Evaluate the integral $\int_1^n \log x \, d\mathcal{L}^1(x)$ and extract the asymptotics]
Integration by parts gives
\begin{align*}
\int_1^n \log x \, d\mathcal{L}^1(x) = \bigl[x \log x - x\bigr]_1^n = n \log n - n + 1.
\end{align*}
Combining the two bounds:
\begin{align*}
n \log n - n + 1 \le \log(n!) \le (n+1)\log(n+1) - n.
\end{align*}
[/step]
[step:Divide by $n \log n$ and squeeze to conclude $\log(n!) \sim n \log n$]
Dividing all three terms by $n \log n$:
\begin{align*}
\frac{n \log n - n + 1}{n \log n} \le \frac{\log(n!)}{n \log n} \le \frac{(n+1)\log(n+1) - n}{n \log n}.
\end{align*}
For the lower bound, $\frac{n \log n - n + 1}{n \log n} = 1 - \frac{1}{\log n} + \frac{1}{n \log n} \to 1$ as $n \to \infty$.
For the upper bound, $\frac{(n+1)\log(n+1)}{n \log n} = \frac{n+1}{n} \cdot \frac{\log(n+1)}{\log n}$. Since $\frac{n+1}{n} \to 1$ and $\frac{\log(n+1)}{\log n} = 1 + \frac{\log(1 + 1/n)}{\log n} \to 1$, the first factor tends to $1$. The remaining term $\frac{-n}{n \log n} = \frac{-1}{\log n} \to 0$. So the upper bound also tends to $1$.
By the squeeze theorem, $\frac{\log(n!)}{n \log n} \to 1$, which is the statement $\log(n!) \sim n \log n$.
[guided]
The asymptotic relation $f(n) \sim g(n)$ means $f(n)/g(n) \to 1$ as $n \to \infty$. We have sandwiched $\log(n!)$ between $n \log n - n + 1$ and $(n+1)\log(n+1) - n$. To verify both bounds give ratio $1$ when divided by $n \log n$:
**Lower bound:** $\frac{n \log n - n + 1}{n \log n} = 1 - \frac{1}{\log n} + \frac{1}{n \log n}$. Both correction terms vanish as $n \to \infty$, so this tends to $1$.
**Upper bound:** We split $\frac{(n+1)\log(n+1) - n}{n \log n} = \frac{(n+1)\log(n+1)}{n \log n} - \frac{1}{\log n}$. The second term vanishes. For the first term, write $\log(n+1) = \log n + \log(1 + 1/n)$, so
\begin{align*}
\frac{(n+1)\log(n+1)}{n \log n} &= \frac{(n+1)}{n} \cdot \frac{\log n + \log(1+1/n)}{\log n} \\
&= \frac{(n+1)}{n}\left(1 + \frac{\log(1+1/n)}{\log n}\right).
\end{align*}
Since $\frac{n+1}{n} \to 1$ and $\log(1 + 1/n) \to 0$ while $\log n \to \infty$, this entire expression tends to $1 \cdot 1 = 1$. Subtracting $\frac{1}{\log n} \to 0$ gives limit $1$.
By the squeeze theorem, $\frac{\log(n!)}{n \log n} \to 1$.
[/guided]
[/step]
