Any suggestions on using induction to prove this inequality? I am solving an exercise and can't advance in the following induction:

$$n\log n - n + 1 \leq \log n!.$$

If necessary, i put the complete question.


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*Update



Calculate
  $$\lim_{n\to \infty}\frac{n!e^{n}}{n^{n}}$$
  following the steps bellow:
A. Show that:
  $$\int\limits_{1}^{n}\log x\,\mathrm{d}x = n\log n - n + 1 = A_{n}.$$
  B. If $B_{n}$ is the right Riemman sum of the function $\log x$
  relative to the partition $\lbrace 1, ..., n\rbrace$ of the interval
  $[1, n]$, show that:
  $$A_{n} \leq B_{n} = \sum_{k = 2}^{n}\log k = \log n!.$$
C.
D.
E.
F.

The steps C, D, E and F are not relevant for my doubt.

Starting from the fundamental $(1+\frac{1}{n})^n \leq e$ for all $n>0$, we get the inequality $$\tag{*} en^n \geq (n+1)^n$$
I'd prefer to work with exponentials over logs, so note that your inequality is equivalent to $$\tag{H} n! \geq e\left(\frac{n}{e}\right)^n $$
For the inductive step, we assume $n! \geq e\left(\frac{n}{e}\right)^n $ and want to show $(n+1)! \geq e\left(\frac{n+1}{e}\right)^{n+1}$.  
This follows as below.  The first inequality is the inductive hypothesis (H) and the second inequality is our knowledge of a lower bound for $e$ (*)
$$(n+1)! = (n+1) n! \geq (n+1) e\left(\frac{n}{e}\right)^n = \frac{n+1}{e^n} en^n \geq \frac{n+1}{e^n}(n+1)^n = e \left(\frac{n+1}{e} \right)^{n+1}$$
Don't forget to establish the base case!
