$n\log (n) − n + 1 \le \log (n!) \le (n + 1) \log (n + 1) − n$ prove using integral

Question

Can someone help me to prove this:

$$n\log (n) − n + 1 \le \log (n!) \le (n + 1) \log (n + 1) − n$$

using integrals? Thank you,

Edit: i tried using that, but i don't understand why

$$\sum_{k=1}^{n}\log k\ge \int_1^{n}\log xdx $$

Thank you

Answer 1

Make a sketch

to see that

$$\int_2^{n+1}f(x-1)dx \le \sum_{k=1}^{n}f(k)\le \int_1^{n+1}f(x)dx$$

Answer 2

Hints:

$$\log(n!)=\log(n)+\log(n-1)+\cdots +\log(2) +\log(1)$$

$$\int \log(x) \, dx = x \log(x)-x +C$$