Help understanding Polya's proof of the Arithmetic-Geometric Mean Inequality Background
Steele, in The Cauchy-Schwarz Master Class, states the general Arithmetic-Geometric Mean Inequality as follows (pg 23): 

$a_{1}^{p_{1}}a_{2}^{p_{2}} \ldots a_{n}^{p_{n}} \leq p_{1}a_{1} + p_{2}a_{2} + \ldots + p_{n}a_{n}$ for non-negative real numbers $a_1,a_2, \ldots a_n$ and each sequence $p_1, p_2, \ldots, p_n$ of positive real numbers which sums to one.

Steele then discusses and presents Polya's "dream" proof which I've summarized below:
Let $A:=p_{1}a_{1} + p_{2}a_{2} + \ldots + p_{n}a_{n}$ and $G:=a_{1}^{p_{1}}a_{2}^{p_{2}} \ldots a_{n}^{p_{n}}$.
We have $1+x\leq e^{x}$ for all $x\in \mathbb{R}$.  By change of variables we have: $x\leq e^{x-1}$ for all $x\in\mathbb{R}.$ 
Then $$\frac{a_{k}}{A}\leq e^{(\frac{a_{k}}{A})-1}$$ for $k \in \{1,2,\ldots, n\}$.
Thus we have $$\left(\frac{a_{1}}{A}\right)^{p_{1}}\left(\frac{a_{2}}{A}\right)^{p_{2}} \ldots \left(\frac{a_{n}}{A}\right)^{p_{n}} \leq \exp{\left(\left\{\sum_{k=1}^{n}p_{k}\frac{a_{k}}{A}\right\}-1\right)}=1.$$
After we move the multiples of $A$ to the righthand side and note that $p_{1}+p_{2}+\ldots + p_{n}=1$, the general Arithmetic-Geometric Mean Inequality is proven.
My Question
I can follow the proof and see that it is correct.  But where do the $\frac{a_{k}}{A}$ terms come from? 
Steele does discuss this but I don't understand his explanation. 
My summary of his explanation is as follows:
If we apply $x\leq e^{x-1}$ when $x=a_{k}$ then we get $a_{k}\leq e^{a_{k}-1}$.  By raising both sides to $p_{k}$ and multiplying, we get $G \leq \exp{(A-1)}$.  
Similarly applying $x\leq e^{x-1}$ with $x=A$ we get $A\leq \exp{(A-1)}$.  
So we have the double bound $$\max{\{A, G\}} \leq \exp{(A-1)}.$$  If either $A$ or $G$ is equal to $\exp{(A-1)}$ then we'll be able to turn this double bound into an inequality between $A$ and $G$, which we want.  Steele says that if we think about trying to exploit this, we're led to the idea of normalization and so we introduce the $\frac{a_{k}}{A}$ terms.  
This last part is what I don't follow.  I'm not sure what Steele means by normalization here or how we're led to this from thinking about the case in which $A$ or $G$ is equal to $\exp{(A-1)}$.  Can anyone help me understand what's going on here?  Thanks!
 A: In my opinion the idea to consider $\frac{a_k}{A}$ comes rather from "backwards working".


*

*We want $G \leq A \Leftrightarrow \frac{G}{A}\leq 1$.

*To bring in the $p_k$ we note that $A^1 = A^{p_1 + \cdots + p_n} = A^{p_1}\cdots A^{p_n}$.

*So, it makes very much sence to consider the product of the quotients $\frac{a_k^{p_k}}{A^{p_k}} = \left( \frac{a_k}{A}\right)^{p_k}$.

A: There's a simple way to get to this from what you have. We have $G\leq\exp(A-1)$. So we would be done if $\exp(A-1)=A$, and that happens precisely when $A=1$. 
This means we have proved generalised AM-GM for the special case when the weighted arithmetic mean is $1$. But any other case can be reduced to this special case by scaling all the numbers, so it must be true in all cases.
A: Maybe the following would be clear:
$$\prod_{k=1}^n\left(\frac{a_k}{A}\right)^{p_k}\leq \prod_{k=1}^n\left(e^{\frac{a_k}{A}-1}\right)^{p_k}=\prod_{k=1}^ne^{\frac{p_ka_k}{A}-p_k}=e^{\sum\limits_{k=1}^n\left(\frac{p_ka_k}{A}-p_k\right)}=e^{1-1}=1.$$
$x=\frac{a_k}{A}$ it's just a substitution in $e^{x-1}\geq x.$
