How to show these inequalities? To show that $\forall n \in \mathbb{N}$
$$n \log{n}-n+1 \leq \log{(n!)} \leq  (n+1) \log{n}-n+1$$
do I have to use induction? Once at the one inequality and then at the other?
Or is there an other way to show this?

$$\log n!=\log 1+\log 2+\cdots +\log n$$ so $$\log n!\le \int_1^{n+1}\ln x\,\mathrm{d}x=(n+1)\ln(n+1)-(n+1)+1=(n+1)\ln(n+1)-n$$ and perhaps you can toy with that to get the right side.
For the other direction, $$\log n!=\log 2+\cdots +\log n\ge \int_1^n \ln x\,\mathrm{d}x=n\ln n-n+1$$
In general for an increasing function, we have $$\int_{a-1}^b f(x)\,\mathrm{d}x\le f(a)+f(a+1)+\cdots +f(b)\le \int_a^{b+1}f(x)\,\mathrm{d}x$$
