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

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

The first term is zero. $$\sum_{m=0}^{\infty} \frac{mx^m}{m!} = \frac{0 x^0}{0!} + \sum_{m=1}^{\infty} \frac{m x^m}{m!} = \sum_{m=1}^{\infty} \frac{x^m}{(m-1)!} = x\sum_{m=1}^{\infty} \frac{x^{m-1}}{(m-1)!} = x\sum_{m=0}^{\infty} \frac{x^m}{m!}.$$
Continuing your line of thought: $$\sum_{m=0}^{\infty} { mx^m\over m!}=\sum_{m=0}^{\infty} {x^m\over (m-1)!}=x\sum_{m=0}^{\infty} {x^{m-1}\over (m-1)!}=x\left(\frac{x^{-1}}{(-1)!}+\sum_{m=1}^{\infty} {x^{m-1}\over (m-1)!}\right)=\\ x\left(0+\sum_{m=0}^{\infty} {x^m\over m!}\right)=x\sum_{m=0}^{\infty} {x^m\over m!}.$$ Note: $$\frac{1}{\Gamma{(0)}}=\frac1{(-1)!}=0$$. Reference: Wikipedia.