Is this alternative representation of $f(x)=xe^x$ as Maclaurin series correct?

Question

Let $f(x)=xe^x$. I know

$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$

and so

$$f(x)=x\sum_{n=0}^\infty \frac{x^n}{n!}=x(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+...=\sum_{n=0}^\infty \frac{x^{n+1}}{n!}$$

But shouldn't this other representation be also correct?

$f'(x)=e^x+xe^x, f''(x)=2e^x+xe^x, f'''(x)=3e^x+xe^x,... \implies f^{(n)}(x)=ne^x+xe^x \implies f^{(n)}(0)=n$

Because I want $f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$ then

$$f(x)=\sum_{n=0}^\infty \frac{n}{n!}x^n$$

If this other representation is also correct, then how $\sum_{n=0}^\infty \frac{n}{n!}x^n=\sum_{n=0}^\infty \frac{x^{n+1}}{n!}$?

Answer 1

The index of the second sum can start from $1$ because $0/0!=0/1=0$. Then, with reindexing, $$\sum_{n=1}^\infty\frac n{n!}x^n=\sum_{n=1}^\infty\frac 1{(n-1)!}x^n=\sum_{n=0}^\infty\frac{x^{n+1}}{n!}$$

Answer 2

Indeed $f^{(n)}(0)=n$. To make the connection, note that in the first case the first term is zero, so you have $\sum_{n=1}^\infty \frac{1}{(n-1)!} x^n$, and now you can shift the index to start at $0$ again by replacing $n$ with $n+1$ everywhere. This gives $\sum_{n=0}^\infty \frac{1}{n!} x^{n+1}$ as you expected.