I found the proof by Kronecker that the expression 

$$X = p^e + p^{e-1} + ... + p^2 +p + 1$$ 

is irreducible.  Four people translated the German, two in Detroit and two in Kiel, Germany.  See 
http://www.oddperfectnumber.org/docs/Kronecker%20proof%20of%20irreducibility.pdf

Kronecker also proved that the numeric factors of $X$ have these forms:  

$(e + 1) | | X,$ and $(k(e + 1) + 1) | X.$

I need the proof of the form of the factors of $\sigma(p^e)$ for a paper on a new result in the odd perfect number problem. I would like the proof in these ways:

1. An exact citation to "Leopold Kronecker's Werke" available on Google Books. 
The article name, the page number, the equation number, etc. to allow me to translate a page or two to understand his proof. 
2. A proof by a reader who would like to share it with us.  There are usually multiple ways to prove any theorem. Please share them all.
3. A citation of an article, preferably in English, showing the proof.

My knowledge of German is to count to 10, my French is a little better, and my Latin went out with Vatican II.  However, I gained some insight into the proof by studying it.

Let $X = \sigma(p^e)  = p^e + p^{e-1} + ... + p^2 +p + 1$, and $p = (j(e + 1) + a)$, where $p$ and $(e + 1)$ are prime. Then 

$$Y = \sigma(p^e)\bmod (e + 1)  = a^e + a^{e-1} + ... + a^2 +a + 1.$$

For $a = 1$, then $Y \bmod (e + 1) = e + 1$.  Removing the factor $(e + 1)$ leaves $1$, not $0$. Therefore, when $p =  j(e + 1) + 1$, then $(e + 1) || X$.

For $a = 0 \bmod (e + 1)$, then $p$ is composite and not a prime.
For $a = 2, 3, ..., e$, then the following holds:

$$Y \bmod (e + 1) = a (a^{e-1} + a^{e-2} + ... + a^2 +a + 1) + 1$$

The complicated term in the parentheses is reducible into several algebraic terms.
Suppose $a = f_1^{e1}  f_2^{e2} ... f_m^{em}$.  Then 

$$Y = a \times \sigma(a^{f_1-1}) \times \sigma(a^{f_2-1}) \times
 ... \times \sigma(a^{f_m-1})  \times
      \sigma((-a)^{f_1-1}) \times \sigma((-a)^{f_2-1}) \times ... \times \sigma((-a)^{f_m-1}) + 1$$, 

I think, maybe, perhaps.  For every value of $a$, one of the terms is zero.  Thus, for $p$ and $a$ as above, $X \bmod (e + 1) = 1$.    One thing for sure, Kronecker proved it better.

Can someone please help me find a proof and the citation? The paper will be scrutinized by the best mathematicians and harshest critics, you.