Getting two very different answers for Maclaurin series for $xe^{x}$ - how is this possible?

$$\sum_{n=0}^\infty \frac{nx^n}{n!} = \sum_{n=1}^\infty \frac{nx^n}{n!} = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}$$

Good question!

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We may see that $$f(x)=\sum_{n\ge0}\frac{n}{n!}x^n=0+\sum_{n\ge1}\frac{n}{n!}x^n.$$ We then see that $n/n!=1/(n-1)!$ when $n\ge1$. Thus $$f(x)=\sum_{n\ge1}\frac{x^n}{(n-1)!}=\sum_{n\ge0}\frac{x^{n+1}}{n!}=xe^x.$$