Why $ \lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} =5e$? I have a small exercise and I don’t know who to get the result.

The exercise is: $$ \lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} $$

I did following transformations:
$$
\frac{(5n^3-3n^2+7)(n+1)^{n-2}}{n^{n+1}} \\
(5n^{2-n}-3n^{1-n}+7^{-1-n})(n+1)^{n-2} \\
(\frac{5}{n^{-2+n}} - \frac{3}{n^{-1+n}} + \frac{7}{n^{1+n}})(n+1)^{n-2}
$$
But none of them helped me to see the result.
It would be great if someone could explain it to me.
Edit
@adrian-barquero
Ok. Fist you factories $^n$ and get
$$
\frac{(n+1)^n}{n^n} = (1+\frac{1}{n})^n = e \\
$$
In the other fraction I could extend with $n^3$
$$
\frac{5n^3-3n^2+7}{n(n + 1)^2} = \frac{n^3(5 - \frac{3n^2}{n^3} + \frac{7}{n^3})}{n^3(1 + \frac{2n^2}{n^3} + \frac{n}{n^3})} = 5
$$

I would recommend rearranging as 
$$
\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} = \frac{5n^3-3n^2+7}{n(n + 1)^2}\frac{(n+1)^n}{n^n} = \frac{5n^3-3n^2+7}{n(n + 1)^2} \left ( 1 + \frac{1}{n}  \right )^n
$$
