Bounds for n! by comparing with an integral

Antiderivative. $\int\ln x\,dx=x\ln x-x$, so $\int_1^n\ln x\,dx=n\ln n-n+1$.

Lower bound. Since $\ln$ is increasing, $\ln k\ge\int_{k-1}^k\ln x\,dx$ for $k\ge2$. Summing from $k=2$ to $n$:
$$\ln n!\ge\int_1^n\ln x\,dx=n\ln n-n+1.$$

Upper bound. Since $\ln$ is increasing, $\ln k\le\int_k^{k+1}\ln x\,dx$. Summing from $k=1$ to $n-1$:
$$\ln(n-1)!\le\int_1^n\ln x\,dx=n\ln n-n+1,$$
so $\ln n!\le n\ln n-n+1+\ln n$.

Exponentiating. $n\ln n-n=\ln\left(\frac ne\right)^n$, which gives exactly the two stated inequalities.

Check. For $n=10$: $e\left(\frac{10}e\right)^{10}\approx1.23\cdot10^6\le10!=3\,628\,800\le1.23\cdot10^7$.

Remark. Stirling's formula $n!\sim\sqrt{2\pi n}\,(n/e)^n$ refines this: the true size lies between the two bounds, around $\sqrt n$ times $(n/e)^n$.