Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$

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e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in N\rbrace
where n=p_r^{e_r}


I am using it in a proof (we were actually shown how to do the proof a different way, but want to see if I can complete the proof in a different way. What I have stated seems logical to me, but not sure how I would prove it. I'm not 100% on the notation I've used, so if anything is unclear I can write what I mean in words. Thanks :)

hmmm, \hspace doesn't seem to work in this environment :S
 
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if 0 \in N then I get e_r+1. If N starts with 1, then OK.

Example: How many powers of 2 divide 8 = 2^3 ... Answer: 4 of them, namely 1,2,4,8.
 
The problem is not the first element of \mathbb N to be 0 or 1; rather, is that 1 is always a divisor of any integer, so the set will always have e_r + 1 elements.
 
I think g_edgar is right. If 0\notin \mathbb N then the set does not contain 1.

We were given

\Lambda(n) := \left\lbrace\begin{array}{cc}<br /> \log p &amp; \textup{if } n \textup{ is a power of a prime } p\\<br /> 0 &amp; \textup{if } n=1 \textup{ or } n \textup{ is a composite number}<br /> \end{array}
Prove that \Lambda(n)= \sum_{d\mid n}{\mu\left(\frac{n}{d}\right) \log d}
Hint: Calculate \sum_{d\mid n}{\Lambda (d)} and apply the Mobius Inversion Formula.
I don't think the definition really makes sense, as I thought a composite number to include powers of primes, but it seems assumed that it means 0 OTHERWISE. Using the property that I gave in the first post, I got:

\sum_{d\mid n}{\Lambda (d)}=\log n

which is what we were given in class when shown how to do this differently, so I figured it must make sense. Obviously because the powers of p do not begin with e_1=0,\;\;\;k\in\mathbb N\;\;\;that I have given, does not start at 0, but 1. Am I making sense? If I am right, how would I prove it? ('It' being the property that I gave in the first post. If I can prove that, then the rest of the problem works itself out.)
 
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To make the first statement in my last post clearer...

1\notin \mathbb P
\therefore d=1\Leftrightarrow k=0EDIT:
So I guess I should write
e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in \mathbb N^*\rbrace
where n=p_r^{e_r}
I always think of the natural numbers as starting at one, but I guess it is optional and one should be explicit about what the set \mathbb N contains. So if someone can help me figure out how to prove the statement, I would really appreciate it.
 
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