Finding the sum $\sum_{j=k}^n (-1)^{j+k}\binom{n}{j}\binom{j}{k}$ I am prepping for my mid semester exam, and came across with the following question:

Find the closed form for the sum $\sum_{j=k}^n (-1)^{j+k}\binom{n}{j}\binom{j}{k}$, using the assumption that $k = 0, 1,...n$ and $n$ can be any natural number.

So what I have done is to note the fact that $$\binom{n}{j}\binom{j}{k}= \frac{n!}{j!(n-j)!}\frac{j!}{k!(j-k)!}=\frac{n!}{(n-j)!\ k!\ (j-k)!}$$
Then we can write the summation as $$\sum_{j=k}^n {(-1)^{j+k}\binom{n}{j}\binom{j}{k}}= \sum_{j=k}^n (-1)^{j+k} \frac{n!}{(n-j)!\ k!\ (j-k)!} = \frac{n!}{k!} \sum_{j=k}^n (-1)^{j+k} \frac{1}{(n-j)!\ (j-k)!} $$
I tried to let $m=j-k$:
$$\frac{n!}{k!} \sum_{m=0}^{n-k} (-1)^{m+2k} \frac{1}{(n-m-k)!\ m!}=\frac{n!}{k!} \sum_{m=0}^{n-k} (-1)^{m} \frac{1}{(n-m-k)!\ m!}$$
But I am not sure what to proceed next. Any help would be highly appreciated!

Note that 
$$\frac{1}{(n-m-k)!\ m!}=\frac{1}{(n-k)!}\cdot \frac{(n-k)!}{(n-m-k)!\ m!}=\frac{1}{(n-k)!}\binom{n-k}{m}$$
Then, you'll have
$$\frac{n!}{k!\ (n-k)!}\sum_{m=0}^{n-k}(-1)^m\cdot 1^{n-k-m}\cdot\binom{n-k}{m}$$
