I am trying to evaluate
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}(1+x+x^2+...+x^n)^d\Bigr|_{x=0}.$$
If one starts deriving directly (or after summing the geometric series) the number of terms gets doubled every time you derive. My approach was the following. We can write
$$(1+x+x^2+...+x^n)^d=c_0+c_1z+c_2z^2+...+c_{n\cdot d}z^{n\cdot d}.$$
The $n$th derivative of this will have the form
$$c'_n+c'_{n+1}z^{n+1}+...+c'_{n\cdot d}z^{n\cdot d},$$
so evaluating at $z=0$ is the same as computing $c'_n$, which
$$c'_n=c_n\cdot n\cdot(n-1)\cdot...\cdot2=c_n\cdot n!.$$
Using the multinomial theorem we have
$$(1+x+x^2+...+x^n)^d=\sum_{k_0+k_1+...+k_n=d}{d \choose k_0,k_1,...,k_n}\prod_{t=0}^nz^{t\cdot k_t}.$$ Now
$$\prod_{t=0}^nz^{t\cdot k_t} = z^{\sum_{t=0}^nt\cdot k_t},$$
so we need $k_t$ such that $\sum_{t=0}^nt\cdot k_t=n$, or $\sum_{t=1}^nt\cdot k_t=n$ and $k_0=d-(k_1+...+k_n)$. Then, $c_n$ will be the the multinomial coefficient where
$$k_1+2k_2+3k_3+...+nk_n=n.$$
I am stuck here, since there are many combinations of $k_t$ that satisfy the previous equality. Any help on how to proceed, or a different way to approach this will be appreciated.
This $n$-th derivative is $n!$ times the $x^n$-coefficient of $$f(x)=(1+x+x^2+\cdots+x^n)^d.$$ But compare this coefficient to that of $$g(x)=(1+x+x^2+\cdots+x^n+x^{n+1}+\cdots)^d=\frac1{(1-x)^d}.$$ These coefficients are the same, since the extra terms don't affect the powers of $x$ up to $x^n$. But, by the binomial theorem, say, $$g(x)=(1-x)^{-d}=\sum_{n=0}^\infty\binom{n+d-1}{d-1}x^n.$$ Your derivative is thus $$n!\binom{n+d-1}{d-1}=\frac{(n+d-1)!}{(d-1)!}$$