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 $
S_n = \left\{ {1,2,...,n} \right\}
$
there are $
\left\lfloor {\tfrac{n}
{p}} \right\rfloor
$
multiples of $p$. But $
\left\lfloor {\tfrac{n}
{{p^2 }}} \right\rfloor
$
of these, are multiple of $
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 $
\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 $
S_n = \left\{ {1,2,...,n} \right\}
$
there are $
\left\lfloor {\tfrac{n}
{p}} \right\rfloor
$
multiples of $p$. But $
\left\lfloor {\tfrac{n}
{{p^2 }}} \right\rfloor
$
of these, are multiple of $
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 $
\left\lfloor {\tfrac{n}
{p}} \right\rfloor + \left\lfloor {\tfrac{n}
{{p^2 }}} \right\rfloor + ...
$
I am still lost in this proof.