Is there a simple way to prove this identity using double summation? I am trying to prove that $$\sum_{k=0}^n\sum_{r=0}^k{n \choose k}{k \choose r}(-1)^{n+k}A_r=\sum_{r=0}^n\sum_{k=r}^n{n \choose k}{k \choose r}(-1)^{n+k}A_r$$
By writing out the full sum I can see that this is true, but is there a method based on purely some properties of double summation?

The nature of the summand doesn't actually matter here: one can show that
$$ \sum_{k=0}^n \sum_{r=0}^k a_{k,r} = \sum_{r=0}^n \sum_{k=r}^n a_{k,r}. \tag{1} $$
Why? We're not summing over the whole square $0\leqslant k,r\leqslant n$, but instead, the just the triangle where $r \leqslant k$:
$$ \sum_{ \substack{0 \leqslant k,r \leqslant n \\ r \leqslant k }} $$
This means that if we swap the order of summation (and sum "vertically" instead of "horizontally", we have to start $k$ at $r$ and sum up to $n$, then do the sum over $r$, so you end up with (1).
