How to find the closed form of the following generating function? $$\sum_{m\geq0}\big[\sum_{k=0}^n \binom{k}{m}\big]z^m
$$
It is supposed to be $$\frac{(x+1)^{n+1}-1}{x}$$ according to Mathematica.
How is it solved?

Rearrange across terms so you are considering a fixed $k$ rather than a fixed $m$ in the secondary sum (can be done, because we find that actually only finitely many terms are non-zero). 
You end up getting, for each $k$, the sum 
$$
\sum_{m \geq 0} \binom km x^m.
$$
But $\binom km = 0$ whenever $m > k$. So you get 
$$
\sum_{m = 0}^k \binom km x^m
$$
which is just the binomial expansion of $(x + 1)^k$.
So your whole sum becomes
$$
\sum_{k = 0}^n (x + 1)^k
$$
but this is just the truncated power series, which satisfies
$$
(x+1)^{n+1} - 1 = ((x+1) - 1) \sum_{k = 0}^n (x+1)^k
$$
So your sum is equal to 
$$
\frac{(x+1)^{n+1} - 1}{(x+1) - 1} = \frac{(x+1)^{n+1} - 1}{x}
$$
