Question about product of generating functions in a proof that for positive integer $n$, $\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1$. I was recently looking at a copy of the Mathematics Magazine from 2004 and was reading Q944 (here). It asks this: 

Show that for positive integer $n$, $$\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1.$$

The solution is here. Basically, if we let $S_n$ be the sum, we find that we can write $$S_n=\sum_{k=0}^na_{n-k}b_k,$$ where $a_k=(-1)^k\binom nk$ and $b_k=\binom{n+k}n$. Then we can find generating functions for $a_k$ and $b_k$. In particular, we find that $$\sum_{k=0}^na_kx^k=(1-x)^n$$ and $$\sum_{k=0}^\infty b_kx^k=\frac1{(1-x)^{n+1}}.$$ 
So far, all of this makes sense to me. Now for the final step of the solution, we note that $$\sum_{n=0}^\infty S_nx^n=\sum_{n=0}^\infty\left(\sum_{k=0}^na_{n-k}b_k\right)x^n=(1-x)^n\cdot\frac1{(1-x)^{n+1}}.$$ 
This last step doesn't really make a lot of sense to me. After all, aren't the generating functions for $a_k$ and $b_k$ dependent on $n$? And so, for example, don't we get that $a_1$ means different things depending on what $n$ is? 
Sorry if I'm not being clear here--I'm having a bit of trouble formulating exactly what my confusion is. But, basically, if somebody could explain the last step in a bit more detail, that'd be fantastic. 
 A: I’d carry out the last step a bit differently. What the first part shows is that $S_n$ is the coefficient of $x^n$ in the product
$$(1-x)^n\cdot\frac1{(1-x)^{n+1}}\;,\tag{1}$$
something that is often written
$$S_n=[x^n]\left((1-x)^n\cdot\frac1{(1-x)^{n+1}}\right)$$
with the $[x^n]$ operator. Clearly, then,
$$\begin{align*}
S_n&=[x^n]\left(\frac{(1-x)^n}{(1-x)^{n+1}}\right)\\
&=[x^n]\left(\frac1{1-x}\right)\\
&=[x^n]\sum_{k\ge 0}x^k\\
&=1\;.
\end{align*}$$
The $n$ in $(1)$ really does depend on which $S_n$ we’re computing, but $(1)$ simplifies to $\frac1{1-x}$ for all $n$, so in the end we really are looking at one power series.
A: It might also help to make the two meanings of $n$ explicit by using different letter for the sum index. For arbitrary but fixed integer $n$, we have generating functions $$f(x)=\sum_{k=0}^\infty a_kx^k=(1-x)^n, g(x)=\sum_{k=0}^\infty b_kx^k=\frac1{(1-x)^{n+1}},$$
with
$$f(x)g(x)=\sum_{m=0}^\infty\left(\sum_{k=0}^ma_{k}b_{m-k}\right)x^m=(1-x)^n\cdot\frac1{(1-x)^{n+1}}=\frac{1}{1-x}.$$
Now to be precise, we should compare coefficients of $x^n$ on both sides, then the intended identity follows. But the right side does not depend on $n$ anymore, which is partly what causes the confusion. However following slightly modified example shows why that it is important to look at that specific coefficient.
Instead of the original problem, consider $a_k=\binom{n}{k}$,$b_k=\binom{n}k$ (again for arbitrary but fixed integer $n$), then actually $f(x)=\sum_{k=0}^\infty a_kx^k=(1+x)^n$, $g(x)=\sum_{k=0}^\infty b_kx^k=(1+x)^n$, and so
$$
f(x)g(x)=(1+x)^n(1+x)^n=(1+x)^{2n}.
$$
Now comparing coefficients at $x^m$ on both sides, we see
$$
\sum_{k=0}^m \binom nk \binom{n}{m-k}=\binom{2n}{m}.
$$
If our goal now was to prove identity $\sum_{k=0}^n \binom nk \binom{n}{n-k}=\binom{2n}{n}$, we would have to look at coefficient of $x^n$, other coefficients would not help indeed (even though they give more general identity, but you get the point).
Notice that in both examples, $n$ is fixed integer, it is not used as an index in any sum, which hopefully helps to see through the argument.
A: $$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{2n-k}{n}
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{2n-k}{n-k}\tag1\\
&=\sum_{k=0}^n(-1)^n\binom{n}{k}\binom{-n-1}{n-k}\tag2\\
&=(-1)^n\binom{-1}{n}\tag3\\[9pt]
&=1\tag4
\end{align}
$$
Explanation:
$(1)$: symmetry of Pascal's Triangle: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: negative binomial coefficients
$(3)$: Vandermonde Identity
$(4)$: $\binom{-1}{n}=(-1)^n\binom{n}{n}$ (negative binomial coefficients)
The equation in step $(2)$ is very close to what you are looking at: the convolution of the coefficients for $(1-x)^n$ and $(1-x)^{-n-1}$. Vandermonde's Identity is based on just this sort of product and gives the coefficients for $(1-x)^{-1}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[5px,#ffd]{\left.\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{2n - k \choose n}
\right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 0}}} 
\\[5mm] = &\
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{z^{n}}\pars{1 + z}^{2n - k}
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{2n}\sum_{k = 0}^{n}
{n \choose k}
\pars{-\,{1 \over 1 + z}}^{k}
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{2n}
\pars{1 - {1 \over 1 + z}}^{n}
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{n}z^{n} = \bbx{\large 1} \\ &
\end{align}
