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.

Hint:
$$
\begin{align}
&((n+1)\log(n+1)-(n+1)+1)-(n\log(n)-n+1)\\
&=\log(n+1)+n\log\left(1+\frac1n\right)-1\\
\end{align}
$$
and
$$
n\log\left(1+\frac1n\right)=\int_n^{n+1}\frac nx\,\mathrm{d}x\lt1
$$
