I have been going through Hanson's proof (page 33-37) and I was following it up until Lemma 3.
There, Hanson does a magic trick which I am not clear on.
Here is the step:
$$\frac{(n/a_i)^{n/a_i}}{((n-a_i+1)/a_i)^{(n-a_i+1)/a_i}} = \left(1+\frac{1}{(n-a_i+1)/(a_i-1)}\right)^{((n-a_i+1)/(a_i - 1))\times((a_i-1)/a_i)}\left(\frac{n}{a_i}\right)^{(a_i-1)/a_i}$$
If anyone could help me understand how these values are equal, I would appreciate it.
Edit 1:
With help, I was able to figure it out. Here's what I came up with.
$\dfrac{(n/a_i)^{n/a_i}}{((n-a_i+1)/a_i)^{(n-a_i+1)/a_i}} = \dfrac{(n/a_i)^{(n-a_i+1)/a_i}}{((n-a_i+1)/a_i)^{(n-a_i+1)/a_i}}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$
$\left(\dfrac{(n/a_i)}{((n-a_i +1)/a_i)}\right)^{((n-a_i+1)/(a_i-1))\times((a_i-1)/a_i)}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$
$\left(\dfrac{n}{n-a_i +1}\right)^{((n-a_i+1)/(a_i-1))\times((a_i-1)/a_i)}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$
$\left(\dfrac{(n-a_i+1) + (a_i - 1)}{n-a_i +1}\right)^{((n-a_i+1)/(a_i-1))\times((a_i-1)/a_i)}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$
$\left(\dfrac{(n-a_i+1)/(a_i-1) + 1}{(n-a_i +1)/(a_i-1)}\right)^{((n-a_i+1)/(a_i-1))\times((a_i-1)/a_i)}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$
$\left(1 + \dfrac{1}{(n-a_i +1)/(a_i-1)}\right)^{((n-a_i+1)/(a_i-1))\times((a_i-1)/a_i)}\left(\dfrac{n}{a_i}\right)^{(a_i-1)/a_i}=$