Evaluate ord_p(p^N!) Homework Statement

[slamminsammya]

Homework Statement

The question is from Koblitz's "P-adic Numbers, p-adic Analysis, and Zeta Functions". It asks me to prove that
$$ord_p(p^N!)= \sum _{i=1}^{N-1}p^i$$.

Homework Equations

Firstly it would be important to define $$ord_px$$. This is defined to be the exponent of the highest power of the prime p that divides x. The only additional useful bit of information is that, in general,

$$ord_p(ab)=ord_pa+ord_pb$$.

The Attempt at a Solution

So I have an evaluation of the ord function, but I can't show that it equals
$$\sum _{i=1}^{N-1}p^i$$.

Since we have the logarithmic like property for the order function, I can say that

$$ord_p(p^N!)=\sum _{i=1}^{p^N}ord_p(i)$$.

My basic approach then is combinatorial. Firstly, I ask how many numbers x in the interval $$1\leq x\leq p^N$$ have order n. The first such number will be $$p^n$$, and then all other numbers will be multiples of this number, so the next question is to ask how many multiples we can have of $$p^n$$ such that the resulting product is still less than or equal to p^N.

There are going to $$p^{N-n}$$ such coefficients, but out of these, some of them will result in a number whose order is larger than n. These will be precisely those coefficients that are some power of p. There are N-n of these, so in all we have:
$$(p^{N-n})-(N-n)=p^{N-n}+n-N$$ numbers whose order is n.

Now the amount that these numbers contributes to the overall sum will be $$n(p^{N-n}+n-N)$$, and the possible degrees range from 1 to N, so we have:
$$ord_p(p^N!)=\sum _{i=1}^{N}ip^{N-i}+i^2-iN$$.

My trouble is that I cannot see how this can be shown to be equal to the answer given in the book, which is the quite simple $$\sum _{i=1}^{N-1}p^i$$.

 

[Office_Shredder]

You can kick double counting in the teeth with this problem and not even worry about it.

For all the numbers between 1 and p^N, there are p^N-1 of them that are divisible by p (every pth one). There are p^N-2 of them that are divisible by p². If we just add those together, we haven't double counted the p² terms because they were counted once when we counted the terms divisible by p, and once when we counted the terms divisible by p², and they were supposed to be counted twice to begin with

You can probably see how this argument extends

 

[slamminsammya]

Thanks!