Factorizing the factorial
Let n be a natural number; let p be a prime;
m = the exponent of p in the prime factorization of n! ;
s = the sum of the (p-ary) digits of n’s representation in base p.
Theorem m = ( n - s ) / ( p - 1 )
Because of the usual recursive definition of n!, an inductive proof over n seems indicated. Since for n = 0, m = 0 ∧ s = 0, the theorem holds for n = 0.
Consider now the transition from n to n + 1; let k be the exponent of p in the prime factorization of n + 1.
Because ( n + 1 )! = ( n + 1 ) ∙ n! and p is prime
(0) Δ m = k .
Because ( n + 1 ) - n = 1
(1) Δ n = 1 .
Because the p-ary representation of ( n + 1 ) ends on exactly k zeros
(2) Δ s = 1 - k∙( p - 1 )
Combining (1) and (2) yields
(3) Δ (( n - s ) / ( p - 1 )) = k .
Combining (0) and (3) yields the induction step, and thus the proof is completed.