Here we derive a more general identity:
$$ \sum_{j=0}^{m} \binom{m}{j}(x+y)^j \sum_{k=0}^{n} \binom{n}{k} x^k (m-j+n-k)!
= \sum_{j=0}^{m} \binom{m}{j} y^{m-j} (j+n)! \sum_{i=0}^{j+n} \frac{x^i}{i!}. \tag{*} $$
The proof is fairly simple and relies on the following identity:
$$ \int_{0}^{\infty} (t+x)^n e^{-t} \, \mathrm{d}t = n!\sum_{i=0}^{n} \frac{x^i}{i!}. $$
The above identity can be proved either by the mathematical induction on $n$ or using the Poisson process. Then
\begin{align*}
\text{[LHS of (*)]}
&= \sum_{j=0}^{m} \binom{m}{j}(x+y)^j \sum_{k=0}^{n} \binom{n}{k} x^k \int_{0}^{\infty} t^{m-j+n-k}e^{-t} \, \mathrm{d}t \\
&= \int_{0}^{\infty} (t+x+y)^m (t+x)^n e^{-t} \, \mathrm{d}t \\
&= \sum_{j=0}^{n} \binom{m}{j} y^{m-j} \int_{0}^{\infty} (t+x)^{j+n} e^{-t} \, \mathrm{d}t \\
&= \sum_{j=0}^{n} \binom{m}{j} y^{m-j} (j+n)! \sum_{i=0}^{j+n} \frac{x^i}{i!} \\
&= \text{[RHS of (*)]}.
\end{align*}
Ok, I think I've found an answer along with a proof. I will go through a series of claims (of course there may be more efficient ways of going about this). The main result is in Claim 3.
Claim 1. For nonnegative integers $m, r, t$ we have
$$ \sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} = (-1)^{m+t}\binom{r}{m}. \tag*{$(1)$} $$
Proof. We proceed by Egorychev's method. We will use the fact that
$$ \binom{j+r}{t+r} = \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz, $$
where the integral is a contour integral over a circle $|z|=\varepsilon$ for some $0<\varepsilon<\infty$.
Using this fact, and interchanging the summation and integral signs when necessary, we have
\begin{align*}
\sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} &= \sum_{j=0}^{m+t} \binom{m+t}{j} \left( \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz \right) (-1)^{j} \\[1.2ex]
&= \frac{1}{2\pi i}\oint \; \sum_{j=0}^{m+t} \binom{m+t}{j} \frac{(1+z)^{j+r}}{z^{t+r+1}} (-1)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (1+z)^{j} (-1)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (-1-z)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (1 - 1 - z)^{m+t} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (-z)^{m+t} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{r-m+1}} (-1)^{m+t} \, dz \\[1.2ex]
&= (-1)^{m+t}\binom{r}{r-m} \\[1.2ex]
&= (-1)^{m+t}\binom{r}{m}.
\end{align*}
$$\tag*{$\blacksquare$}$$
Claim 2. For nonnegative integers $n, r, s$ with $s\ge r$, we have
$$ \sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{r}{n-s}. \tag*{$(2)$} $$
Proof. If $s > n$, then both sides of $(2)$ are clearly $0$, so for the proof let us assume $s\le n$. Take $m = n-s$ and $t = s-r$ and plug these into $(1)$. We obtain
$$ \sum_{j=0}^{n-r} \binom{n-r}{j}\binom{j+r}{s}(-1)^{j} = (-1)^{n-r}\binom{r}{n-s}. $$
We can shift the summation index by taking $j = k-r$ (where $k$ is the new index). By doing this and then multiplying both sides by $(-1)^{r}$, we get $(2)$.
$$\tag*{$\blacksquare$}$$
Claim 3. For nonnegative integers $n, r, s$, we have
$$ \boxed{ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} } $$
where the RHS involves the multinomial coefficient.
Proof. Without loss of generality, assume $s\ge r$. Consider the fact that
$$ \binom{n}{k}\binom{k}{r} = \binom{n}{r}\binom{n-r}{k-r} $$
and use this to obtain
$$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = \binom{n}{r}\sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k}. $$
By Claim 2, we see that the RHS reduces to
$$ \binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s}. $$
This is then decomposed as
\begin{align*}
\binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s} &= (-1)^{n} \frac{n!}{r!(n-r)!}\frac{r!}{(n-s)!(r-n+s)!} \\
&= (-1)^{n}\frac{n!}{(n-r)!(n-s)!(r+s-n)!},
\end{align*}
and the fraction in the last expression can be identified as a multinomial coefficient, giving us
$$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} $$
as desired.
$$\tag*{$\blacksquare$}$$
Best Answer
Suppose that you want to count the $i$-element subsets of $[i]=\{1,2,\ldots,i\}$. Of course there’s only one of them, but we can also count them by the following roundabout procedure. We first expand the set from which we’re drawing the $i$-element subset to $[i+j]=\{1,\ldots,i+j\}$. Now for each $\ell\in[i]$ let $A_\ell$ be the family of $i$-element subsets of $[i+j]$ that do not contain $\ell$; $\bigcup_{\ell=1}^iA_\ell$ is the family of $i$-elements subsets of $[i+j]$ that are not subsets of $[i]$. By the inclusion-exclusion principle we have
$$\begin{align*} \left|\bigcup_{\ell=1}^iA_\ell\right|&=\sum_{\varnothing\ne I\subseteq[i]}(-1)^{|I|+1}\left|\bigcap_{\ell\in I}A_\ell\right|\\ &=\sum_{k=1}^i\binom{i}k(-1)^{k+1}\binom{i+j-k}i\;, \end{align*}$$
since each non-empty $I\subseteq[i]$ has cardinality in $[i]$, for each $k\in[i]$ there are $\binom{i}k$ subsets of $[i]$ of cardinality $k$, and if $|I|=k$,
$$\left|\bigcap_{\ell\in I}A_\ell\right|=\binom{i+j-k}i\;.$$
There are $\binom{i+j}i$ $i$-element subsets of $[i+j]$ altogether, so after we throw out the ones not contained in $[i]$, we have left
$$\begin{align*} \binom{i+j}i&-\sum_{k=1}^i\binom{i}k(-1)^{k+1}\binom{i+j-k}i\\ &=\binom{i+j}i+\sum_{k=1}^i\binom{i}k(-1)^k\binom{i+j-k}i\\ &=\sum_{k\ge 0}\binom{i}k(-1)^k\binom{i+j-k}i\;, \end{align*}$$
and we already know that this is $1$.
Note that there is no need to specify an upper limit on the summation: $\binom{i}k=0$ when $k>i$, and $\binom{i+j-k}i=0$ when $k>j$, so all terms with $k>i\land j$ are $0$ anyway.