Proof
(related to Proposition: Calculating the Sum of Divisors)
 Let $n > 1$ and let $n=\prod_{p\mid n}p^l$ be the factorization of $n.$
 For each prime number $p$ which is a divisor of $n$ we have thet $1=p^0,p^2,\ldots,p^l$ are all divisors of $n.$
 Therefore, in particular these divisors contribute to the sum of all divisors $F(n)$ and, by the formula for the sum of geometric progression, we have
$$\sum_{m=0}^l p^m=\frac{p^{l+1}1}{p1}.$$
 Now, not only one but any other prime number dividing $n$ contributes this factor.
 By the fundamental counting principle it follows that $F(n)=\prod_{p\mid n}\frac{p^{l+1}1}{p1}.$
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927