Help finding a proof of $\sum_{k=0}^{n}(-1)^{k}\binom{j}{k}=(-1)^{n}\binom{j-1}{n}$
I know it's easy to sum up the left side of the identity when the upper limit is $j$ or $j-1$.
However I have no idea when I need to sum up to an arbitrary limit $n$.
If you really want a combinatorial proof, see Identity 168 in Proofs That Really Count, where the proof uses an almost one-to-one correspondence between even and odd subsets of $\{1,\dots,n\}$ via a mapping that either adds to or removes element $n$ from the subset.
For a generating function proof, it is convenient to reverse the order of summation to obtain the equivalent identity $$\sum_{k=0}^n (-1)^k \binom{j}{n-k} = \binom{j-1}{n} \tag1$$ Now \begin{align} \sum_{n=0}^\infty \sum_{k=0}^n (-1)^k \binom{j}{n-k} z^n &=\sum_{k=0}^\infty (-z)^k \sum_{n=k}^\infty \binom{j}{n-k} z^{n-k} \\ &=\sum_{k=0}^\infty (-z)^k \sum_{n=0}^\infty \binom{j}{n} z^n \\ &=\sum_{k=0}^\infty (-z)^k (1+z)^j \\ &=\frac{1}{1+z} (1+z)^j \\ &=(1+z)^{j-1} \\ &= \sum_{n=0}^\infty \binom{j-1}{n} z^n, \end{align} which immediately implies $(1)$.