So I've done this to a series $$ \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!} \int_0^xt^{n-2} = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!}x^{n-2} $$ But looking at the solution it says $$ \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!} \int_0^xt^{n-2} = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}(n-1)n!}x^{n-1} = \sum_{n=1}^{\infty} \frac{1}{3^{n}(n+1)!n}x^{n}$$
So my question is how can I simplify my solution to obtain it in that way, because I don't understand where does the $ (n-1)n! $ comes from
Note that $$\int_0^x t^{n-2} dt = \frac{t^{n-1}}{n-1} \bigg | _0^x = \frac{x^{n-1}}{n-1} \ne x^{n-2} $$