Proof
(related to Proposition: Difference Operator of Powers)
This follows immediately from the binomial theorem for $(x+y)^n= \sum_{k=0}^n{n\choose k}x^{n-k}y^k$ for $y=1.$
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References
Bibliography
- Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960
