<s>[INST] coefficient on $s^{14}$ in generating function I have $(s+s^2+s^3+s^4+s^5+s^6)^7$, and I'm trying to find the coefficient on $s^{14}$. I've tried using the multinomial theorem, but that leads to the problem of finding all $k_1, k_2, \ldots , k_6$ such that $\sum_{n=1}^6 k_n = 7$ and $\sum_{n=1}^6nk_n = 14$, and that doesn't seem to put me any closer to an answer.
I tried rewriting it as $s^7(1+s+s^2+s^3+s^4+s^5)^7$ and looking for the coefficient of $s^7$ in the right half, $(1+s+s^2+s^3+s^4+s^5)^7$, but that doesn't make things much easier.
 [/INST] Binomial Series Approach
Using the Binomial Theorem and negative binomial coefficients,
$$
\begin{align}
\left(s+s^2+s^3+s^4+s^5+s^6\right)^7
&=s^7\left(\frac{1-s^6}{1-s}\right)^7\\
&=s^7\sum_{k=0}^7\binom{7}{k}\left(-s^6\right)^k\sum_{j=0}^\infty\binom{-7}{j}(-s)^j\\
&=s^7\sum_{k=0}^7(-1)^k\binom{7}{k}s^{6k}\sum_{j=0}^\infty\binom{j+6}{j}s^j\tag1
\end{align}
$$
Using Cauchy Products, the coefficient of $s^n$ in $(1)$ is
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}\tag2
$$
Plug $n=14$ into $(2)$:
$$
\bbox[5px,border:2px solid #C0A000]{\binom{7}{0}\binom{13}{7}-\binom{7}{1}\binom{7}{1}=1667}\tag3
$$

The Sum in $\bf{(2)}$ represents a Polynomial
When $n\ge43$, $n-6k-1\ge0$ for all $k\le 7$. This means that 
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7} = \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{6}
$$
which is an order $7$ repeated difference of a degree $6$ polynomial in $k$. Therefore, for $n\ge43$,
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}=0
$$
That is, for $n\ge43$, the coefficient of $x^n$ vanishes.
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