Thread: prime factorization

Let with n>1, and let p be a prime number. If n!, prove that the exponent of p in the prime factorization of n! is [n/p] + [n/p^2] + [n/p^3] +......(Note that this sum is finite, since [n/p^m]=0 if p^m>n

Originally Posted by rmpatel5

Let with n>1, and let p be a prime number. If n!, prove that the exponent of p in the prime factorization of n! is [n/p] + [n/p^2] + [n/p^3] +......(Note that this sum is finite, since [n/p^m]=0 if p^m>n

Note that in the set there are multiples of . But of these, are multiple of and so on.

You must count once the numbers which are multiples of p but not of p², twice the numbers which are multiple of p² and not of p³,... , in order to get the maximum power of p dividing n!

And to do so it's enough to sum

Originally Posted by PaulRS

Note that in the set there are multiples of . But of these, are multiple of and so on.

You must count once the numbers which are multiples of p but not of p², twice the numbers which are multiple of p² and not of p³,... , in order to get the maximum power of p dividing n!

And to do so it's enough to sum

I am still lost in this proof.