# 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+

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

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

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

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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