(HN comments aren't the best medium for writing mathematics. I hope this ends up reasonably readable.)
(d choose k) is the number of ways to pick k things from d.
(d choose k) k^2 is the number of ways to pick k things from d, and then colour one of them red and one of them (possibly the same one) blue.
(-1)^k (d choose k) k^2 is that (when k is even), or minus that (when k is odd).
So the identity says: Suppose we consider a set of d things, and we consider picking some of them (meaning 0 or more, though actually 0 is impossible; k is the number we pick) and colouring one of those red and one blue; then there are the same number of ways to do this picking an odd number of things as picking an even number.
Well, let's describe the process a bit differently. We take our set of d things. We colour one of them red. We colour one of them blue. Then we pick some set of the things (k in number), including the red thing and the blue thing (which may or may not also be the red thing). And the claim is that there are the same number of ways to do this with an even value of k as with an odd number.
But now this is almost trivial!
Once we've picked our red and blue things, we just have to decide which of the other things to pick. And as long as there's at least one of those, there are as many ways to pick an odd number of things as to pick an even number. (Fix one of them; we can either include it or not, and switching that changes the parity.)
So if d>2 then we're done -- for each choice of red and blue, there are the same number of "odd" and "even" ways to finish the job.
What if d<=2? Well, then the "theorem" isn't actually true.
>We take our set of d things. We colour one of them red. We colour one of them blue. Then we pick some set of the things (k in number), including the red thing and the blue thing
In this alternate view of the process, where does the k^2 term come from? Isn't the number of ways to do it in this perspective just
(d-1 choose k-1) # We already chose the red one (which is also the blue) and need to choose k-1 more
+
(d-2 choose k-2) # We chose both red and blue ones and need to choose the other k-2
Edit: I think it actually works out, seems to be equal to
d * 1 * Binomial[d - 1, k - 1] + d * (d - 1) * Binomial[d - 2, k - 2] = Binomial[d, k]*k^2
And then you apply your argument about fixing one of the elements in each of the two cases. Actually the proof seems to be a variant (same core argument) as sum of even binomial terms = sum of odd binomial terms [1]
In the alternate view, the k^2 isn't there any more. Not because we are computing a different number, but because we're organizing the computation differently.
(It's turned into a d^2 factor, except that we then need to treat the red=blue case differently from the red!=blue case, so it's really a d factor plus a d^2-d factor.)
And yes, the last bit of the proof is exactly the same thing as one way to prove that the sum of "even" and "odd" binomial coefficients are equal -- provided n>0. (For the (0 choose whatever) case, the "even" sum is 1 and the "odd" sum is 0.)
