Can anyone prove that ∑ (-1)^i * nCi * (k – i)^n = n!?

Prove that
n
∑ (-1)^i * nCi * (k – i)^n = n!
i=0

where i, n, k Є Z+
Update: Sorry that should be
i, n, k Є Z
Update 2: Test it for yourself and see if the result is independent of k.
Update 3: Yes nCi is the combination formula.

For n=1, k=10, LHS = 10^1 - 9^1 = 1
For n=2, k=10, LHS = 10^2 - 2*9^2 + 8^2 = 2
Update 4: For n=0, LHS = 1
For n=3, k= -1, LHS = (-1)^3 - 3*(-2)^3 + 3*(-3)^3 - (-4)^3 = 6
Update 5: Obviously I formulated this question out of the one Zanti referred to, but can anyone do this using induction or some other method that does not derive from successive differences?
Update 6: NO I haven't found a way to do it. If I had, I would have used it to answer the other question.
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