How are you getting the $2p^{15}p^8$ part?

Please see my edit.

I mean, what specifically do you mean by "taking products"? I understand the part about the totient function, but where are you coming up with the expression $p^{15}p^8(p^8-p^7)+p^{15}(p^8-p^7)p^7$? Does this have to do with generators for $\mathbb{Z}_{p^3}$ etc.?

Yeah. I guess you are right, I didn't explain it that lucidly. I just went for the correct answer. Let me try to clarify. I don't need to restrict myself to generators of $\Bbb Z_3,\Bbb Z_5$ and $\Bbb Z_7$, because all of their elements have order dividing $p^8$. So that explains the factor of $p^3p^5p^7=p^{15}$. Now I'm thinking in terms of the lcm property that you mentioned. Do you see where I'm going with this? To get elements of order $p^8$ I look in the last two factors: $\Bbb Z_{p^9}$ and $\Bbb Z_{p^{11}}$. The other factors aren't big enough to have elements of order $p^8$.

Sorry. There were some typos above. I left some $p$'s out. Anyway, both $\Bbb Z_{p^9}$ and $\Bbb Z_{p^{11}}$ have $\varphi(p^8)$ elements of order $p^8$. Are you with me?

Okay, I see the factor of $p^{15}$ now, so where do we get the sum from?

Hi, so are we getting the sum because we can pick a generator from $\mathbb{Z}_9$ or $\mathbb{Z}_11$?

Ok. Here we go. The sum arises because I split it into two parts.

Yes. That's right.

Okay, so my last question is, on each term, where does the factor of $p^8$ (on the first term) and $p^7$ (on the second term) come from?

Do you understand the first term of the sum?

I understand p^{15}(p^8-p^7), but I'm not quite sure about the extra p^8. Namely, how is it instead p^{15}p^8(p^8-p^7)?

Good. For each choice of a generator (element of order p^8) in Z_p^9, I can pair it with anything from the other factors having order dividing p^8.

The p^8 appears because Z_p^{11} has p^8 elements of order dividing p^8.

Remember the lcm property.

And the p^7 appears because Z_p^{9} has p^7 elements of order dividing p^8?

No. Hold on. This is perhaps the trickiest part. And like I said I just went for it. Here it is:

So far I only took generators for the subgroup of order p^8 in Z_p^9. But, there are still the other elements of that subgroup, each having order dividing p^8.

Ok, I am following

How many are there? Well, p^8-(p^8-p^7)=p^7.

And shame on me for leaving that out.

Ah ok, that makes sense now. And no worries, I appreciate you explaining it. Maybe you could add that last part in to your answer about the p^7 so others can see?

Ok. Nice job.

Have a nice day!

You too.