## Finding the radius of convergence of $\sum\frac{(n+1)z^n}{n!}$

I am trying to find the radius of convergence $$R$$ of the complex series $$\sum\frac{(n+1)z^n}{n!}$$. I understand that $$R=\infty$$, but I need a little help understanding one line of the process.

I use the fact that $$R=\lim_{n\to\infty}\left|\frac{a_{n-1}}{a_n}\right|$$. Here, $$a_n=\frac{n+1}{n}$$, so $$a_{n-1}=\frac{n}{(n-1)!}$$. Hence $$\lim_{n\to\infty}\left|\frac{a_{n-1}}{a_n}\right|=\lim_{n\to\infty}\frac{n.n!}{(n+1)(n-1)!}$$. All is well up until this point.

I am struggling with finding this limit. My textbook says $$\lim_{n\to\infty}\frac{n.n!}{(n+1)(n-1)!}=\lim_{n\to\infty}n$$, with no explanation of how that conclusion was reached. I attempted to divide the numerator and denominator by $$n!$$ and got $$\frac{n.n!}{(n+1)(n-1)!}=\frac{n}{\frac{n+1}{n!}\frac{n-1}{n!}}=\frac{n}{\frac{n+1}{n!}\frac{1}{n}}=\frac{n^2.n!}{n+1}$$ – not the result I was looking for.

Any help would be appreciated very much!