1. ## Proof Help

Hi Guys,

I need some help in proving this:

The factorization of n is given as:

$
n = p_{1}^ep_{2}^e...p_{k}^e
$

[the e's in the superscripts are e1, e2 ... ek]

I have to prove that the number of divisors of n is equal to:

$(e_{1} + 1)(e_{2} + 1)...(e_{k} + 1)$

Need help in proving this guys.

2. Note that any factor of $n$ must look like $p^{x_1 } \cdot p^{x_2 } \cdot \cdot \cdot p^{x_k }$.
But also note that $0\le x_j \le e_j$.
Do you see why?

3. Not exactly. Some explanation please.

4. If $0\le n \le 3$ then $n=0,~1,~2,\text{ or }3$that is four values for $n$.

5. Thanks again.

I have some little questions here:

1)
why is it that 1 is added to the powers of the factors?

2)
How do i write the proof as a whole? I mean how do i write it officially?

6. Originally Posted by kap
why is it that 1 is added to the powers of the factors?
Because $x_j=0,~1,~2,\cdots,~e_j$ that is $e_j+1$ values for the exponent.

And no, I will not do this for you.

7. AM not asking you to do it for me but i want you to explain to me how to do it step by step.

thanks

8. Originally Posted by kap
AM not asking you to do it for me but i want you to explain to me how to do it step by step.
That is exactly what I have done.