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MissMoneypenny
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Homework Statement
Let p be a prime, k be positive integer, and m ∈ {1, 2, 3, ..., pk-1}. Without using Lucas' theorem, prove that p divides [itex]\binom{p^k}{m}[/itex].
Homework Equations
The definition of the binomial coefficients: [itex]\binom{a}{b} = \frac{a!}{b! (a-b)!}[/itex]
The Attempt at a Solution
I've managed to prove the statement when k=1 by arguing that p must divide p! but cannot possible divide m!(p-m)!. I figure I should be able to use a similar counting argument for this proof, likely by counting the number of factors of p that appear in the numerator and in the denominator of [itex]\binom{p^k}{m}[/itex]. It looks to me as though the number of factors of p in [itex]p^k ![/itex] is [itex]\sum_{i=1}^{k} p^{k-i}[/itex]. However, I'm having applying this to the problem. If anyone could give some advice on how to do so, or perhaps nudge me in the direction of a different proof it would be much appreciated.
Thanks!
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