Why does index change in this specific sort of sum: $\sum\limits_{m=0}^{\infty} \frac{mx^m}{m!}=x\sum\limits_{m=1}^{\infty}\frac{x^{m-1}}{(m-1)!}$?

Question

I have an equation similar to the one below:

$$ \sum_{m=0}^{\infty} { mx^m\over m!} $$

which naturally reduces to:

$$ \sum_{m=0}^{\infty} {x^m\over (m-1)!} $$

and now for some reason we can do the following:

$$ = x \sum_{m=1}^{\infty} {x^{m-1}\over (m-1)!} $$

Why does the index increase? Is it just because we cannot evaluate $(m-1)!$ if $m=0$?

Note: if this works then we can shift the summation index to get that $ \sum_{m=0}^{\infty} { mx^m\over m!} = xe^x$.

Answer 1

The first term is zero. $$ \sum_{m=0}^{\infty} \frac{mx^m}{m!} = \frac{0 x^0}{0!} + \sum_{m=1}^{\infty} \frac{m x^m}{m!} = \sum_{m=1}^{\infty} \frac{x^m}{(m-1)!} = x\sum_{m=1}^{\infty} \frac{x^{m-1}}{(m-1)!} = x\sum_{m=0}^{\infty} \frac{x^m}{m!}. $$