Exponent becomes factor in bit decomposition multiplication?

Question

Help on the last equivalence will be greatly appreciated.

Why is the $a_i $ from the exponent of $x^{2^i\cdot a_i }$ becomes a factor in $a_i \cdot x^{2^i}$ ? This is from the following paper, page 18.

Many thanks in advance.

Answer 1

Notice that $a_i=0 \text{ or } 1$.

In this case, if $a_i=0$, the coefficients $x^{2^i a_i}=1$ do not influence the product.

If $a_i=1$, you have $x^{2^i a_i}=a_i x^{2^i} +1-a_i$. Then, $$\prod_{i=0}^{l-1} x^{2^i a_i}=\prod_{i=0,\; a_i\neq 0}^{l-1} x^{2^i a_i}=\prod_{i=0,\; a_i=1 }^{l-1} x^{2^i a_i}=\prod_{i=0 }^{l-1} (a_i x^{2^i}+1-a_i).$$