1. ## 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

2. 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 $\displaystyle S_n = \left\{ {1,2,...,n} \right\}$ there are $\displaystyle \left\lfloor {\tfrac{n} {p}} \right\rfloor$ multiples of $\displaystyle p$. But $\displaystyle \left\lfloor {\tfrac{n} {{p^2 }}} \right\rfloor$ of these, are multiple of $\displaystyle p^2$ 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 $\displaystyle \left\lfloor {\tfrac{n} {p}} \right\rfloor + \left\lfloor {\tfrac{n} {{p^2 }}} \right\rfloor + ...$

3. Originally Posted by PaulRS
Note that in the set $\displaystyle S_n = \left\{ {1,2,...,n} \right\}$ there are $\displaystyle \left\lfloor {\tfrac{n} {p}} \right\rfloor$ multiples of $\displaystyle p$. But $\displaystyle \left\lfloor {\tfrac{n} {{p^2 }}} \right\rfloor$ of these, are multiple of $\displaystyle p^2$ 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 $\displaystyle \left\lfloor {\tfrac{n} {p}} \right\rfloor + \left\lfloor {\tfrac{n} {{p^2 }}} \right\rfloor + ...$
I am still lost in this proof.