Reverse Engineering in Number Theory

First you can notice that the expression inside the product is the explicit form for a geometric sum. Therefore we can manipulate the expression a bit:

$\prod _{ i=1 }^{ \infty }{ \left( \frac { 1-{ { p }_{ i } }^{ { e }_{ i }+1 } }{ 1-{ p }_{ i } } \right) } =\prod _{ i=1 }^{ \infty }{ \left( \sum _{ j=0 }^{ { e }_{ i } }{ { { p }_{ i } }^{ j } } \right) }$

An important fact is, that most $e_i$ 's are $0$ . Therefore in the infinite product many of the ${p_i}^{e_i}$ parts will be $1$ and we can ignore them.

Then you could try an example for $n$ (because we are reverse engineering). An Example with only one prime factor: $9$

$\prod _{ i=1 }^{ \infty }{ \left( \sum _{ j=0 }^{ { e }_{ i } }{ { p_i }^{ j } } \right) } =\sum _{ j=0 }^{ 2 }{ { 3 }^{ j } } ={ 3 }^{ 0 }+{ 3 }^{ 1 }+{ 3 }^{ 2 }$

Now you could try an example with two prime factors: $36$

$\prod _{ i=1 }^{ \infty }{ \left( \sum _{ j=0 }^{ { e }_{ i } }{ { p_i }^{ j } } \right) } =\left( \sum _{ j=0 }^{ 2 }{ { 2 }^{ j } } \right) \cdot \left( \sum _{ j=0 }^{ 2 }{ { 3 }^{ j } } \right) =\left( { 2 }^{ 0 }+{ 2 }^{ 1 }+{ 2 }^{ 2 } \right) \cdot \left( { 3 }^{ 0 }+{ 3 }^{ 1 }+{ 3 }^{ 2 } \right)$

If we factor out this product we get some messy stuff. But by rearanging in a grid we get something familiar :

|| $3^2$ || 9 || 18 || 36 ||

And now we have all the divisors of $n$ arranged in a grid (2D-plane). If we would take a number $n$ with 3 prime factors we could rearange these summands in a 3D-space; 4 prime factors in a 4D-space and so on. And we always get the sum of all divisors of our choosen number $n$ .