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)(mod e + 1)  = a^e + a^{e-1} + ... + a^2 +a + 1.

For a = 1, then Y(mod 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(mod e + 1), then p is composite and not a prime.
For a = 2, 3, ..., e, then the following holds:

Y(mod 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 = f1^{e1}  f2^{e2} ... fm^{em}.  Then 

Y = a x sigma(a^{f1-1}) x sigma(a^{f2-1}) x ... x sigma(a^{fm-1})  x 
      sigma((-a)^{f1-1}) x sigma((-a)^{f2-1}) x ... x sigma((-a)^{fm-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(mod 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.