請證明下式

Please see the proof from the image file:
http://img232.imageshack.us/img232/614/proofecm.jpg

http://img232.imageshack.us/img232/614/proofecm.jpg.

OK

PROOF:
In the spirit of Calculus of Differences, or Umbral Calculus.
In the following, let P_n be the set of polynomials of exact degree n .
All operators D which are considered are linear and of type P-->P .
Take the operator D:P-->P having the images (Df)(x)=f(x)-f(x-1). If e_0(x)=1 ,e_1(x)=x ,...,e_n(x)=x^n , then
(De_n)(x)=x^n - (x-1)^n =nx^{n-1}+ ...[a certain polynomial of degree <= n-2]
(D^2 e_n)(x)=n(n-1)x^{n-2}+... [a polynomial of degree <= n-3]
...
(D^k e_n)(x)=n(n-1)...(n-k+1)x^{n-k}+ ...
...

(D_n e_n)(x)= n! .
On the other hand, using induction on n we have (D^n g)(x)=Σ [k=0 to k=n] (-1)^(k)C(n,k)g(x-k) [where C(n,k)=n!/(k!(n-k)!]
Consider g(x)=x^n. We find n! = Σ [k=0 to k=n] (-1)^(k)C(n,k)(x-k)^n
Actually, this is the same identity given by you. Let k=x,x=y
Σ [x=0 to x=n] (-1)^(x)C(n,x)(y-x)^n = n!
If you using forward difference operator E: f(x+1)-f(x), then the identity becomes Σ [x=0 to x=n] (-1)^(x)C(n,x)(y+x)^n = n!