Recall the exponential function's power series,
[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]
The given sum has [tex]n+1[/tex] instead of [tex]n[/tex]. We can adjust the given sum to get something usable. Notice that in the series expansion for [tex]e^x[/tex], the degree of [tex]x[/tex] starts at 0, while in the given series, the degree starts at 1. We can actually rewrite the given series so that it starts at a different index:
[tex]\displaystyle\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}=\sum_{n=1}^\infty\frac{x^n}{n!}[/tex]
Then this is exactly [tex]e^x[/tex] without the [tex]n=0[/tex] term. When [tex]n=0[/tex], we have [tex]\dfrac{x^0}{0!}=1[/tex], so
[tex]\displaystyle\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}=-1+\sum_{n=0}^\infty\frac{x^n}{n!}=e^x-1[/tex]
