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!