Proof that sum of power series equals exponential function? I have found that the Sum series equal an exponential function as below, however I have not found a proof for it:
$$
ze^z = \sum_{k=0}^{\infty} k \frac{z^k}{k!}
$$
I have though managed to prove that the equality holds by Taylor-expanding $$f(z) = ze^z.$$
My question is; how am I supposed to figure out that the summation series equals that exponential function? Is it something that is just supposed to be observed, or is it possible to go backwards, by for example re-writing the terms in the sum?
Thank you !
J
 A: This is equivalent to trying to prove that
$$ 1 = \sum_{k=0}^{\infty} \frac{k}{k!}.$$
You can see that these are equivalent by dividing both sides by $e^z$ (which is never zero). If $z=0$ clearly both sides are the same. If $z\neq 0$, you can divide both sides by $z$ so in either case, you can neglect $ze^z$ on both sides.
Note how similar this is to the exponential function: $e^z = \sum\limits_{n=0}^{\infty} \dfrac{z^n}{n!}$. If you differentiate both sides you get $$e^z = \sum\limits_{n=0}^{\infty} \dfrac{nz^{n-1}}{n!}.$$
What happens if you set $z=1$? Do you get your desired result?
A: One has
$$
 \sum_0\frac{k}{k!} z^k = \sum_1\frac{k}{k!} z^k 
$$
since the $k=0$ factor at the beginning of the summation kills the first term. Now, you have
$$
\sum_1\frac{k}{k!} z^k=\sum_1\frac{1}{(k-1)!} z^k=\sum_0\frac{1}{k!} z^{k+1}
$$
by renumbering the terms in the series (replace $k$ with $k+1$). Note the change in the start of the summation. Next, you have
$$
\sum_0\frac{1}{k!} z^{k+1}=\sum_0\frac{1}{k!} z z^k=z \sum_0\frac{1}{k!} z^k= z \exp(z)
$$
since you can pull out a constant factor $z$ from the summation which is a limit, i.e, you use 
$$\lim  z*S(K)=z*\lim S(K)$$
$S(K)$ being the partial sums over (k=0..k=K). The very last identity in the second to last line involves only the definition of the exponential function Exp.
