Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$

Question

While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ terms, that is, a single formula for the product $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)(x^{a^4}+1)\ldots(x^{a^n}+1)$$

After multiplying the first four terms I could see a pattern develop in the form $x^{a+a^2+a^3+a^4...a^n}+x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}+x^{a+a^2+a^3}+x^{a+a^2+a^4}+x^{a^2+a^3+a^4}+x^{a+a^3+a^4} ...+x^{a^{n-2}+a^{n-1}+a^n} + x^{a+a^2} +x^{a+a^3}+x^{a+a^4}+x^{a^2+a^3}+x^{a^2+a^4}+x^{a^3+a^4}...x^{a^{n-1}+n}...+x^{a+a^2+a^3+....a^{n-2}+a^{n-1}+1}$

Now I can easily find the summation of powers in the first term but cant find a formula for the summation of $x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}$ and also can't figure out how to account for the other terms.

Answer 1

Using $[n]=\{1,2,\ldots,n\}$ we can write the product as \begin{align*} \prod_{j=1}^n\left(x^{a^j}+1\right)=\sum_{S\subseteq [n]}x^{\sum_{j\in S}a^j} \end{align*} where $S$ runs over all subsets of $[n]$. A special case is $S=\emptyset$, which gives the empty sum. The empty sum is zero per definition, resulting in $x^0=1$.

Answer 2

For the specific value of $a=1/2$ (as mentioned in the comments), one has $$\prod_{j=1}^n \left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1},$$ which can be easily proven by induction on $n$: if this holds for $n$, then $$\prod_{j=1}^{n+1}\left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1}\left(x^{2^{-(n+1)}}+1\right)=\frac{x-1}{x^{2^{-(n+1)}}-1},$$ so it holds for $n+1$.