Like the title says, is there a way to simplify this to some polynomial form or any other form that gets rid of the sum and product?
$$\sum_{i=1}^{h}\prod_{k=1}^i(2^{2^{h-k}} + 1)$$
or this:
$$\sum_{i=1}^{h}\prod_{k=1}^i 2^{2^{h-k}}$$
I have no idea how to approach the first one, but the second one I think it might be possible to turn the product into a sum, but not sure. thanks in advance.
I was able to simplify the second summation $\sum_{i=1}^h\prod_{k=1}^i 2^{2^{h-k}}$:
\begin{align*} j&=h-k\\ k&=1\Rightarrow j=h-1\\ k&=i\Rightarrow j=h-i\\ \sum_{k=1}^i 2^{h-k} &= \sum_{j=h-i}^{h-1} 2^j\\ &= \sum_{j=0}^{h-1} 2^j - \sum_{j=0}^{h-i-1} 2^j\\ &= \frac{2^h-1}{2-1} - \frac{2^{h-i}-1}{2-1}\\ &= 2^h-2^{h-i}\\ \sum _{k=1}^i 2^{h-k} &= 2^h-2^{h-i}\\ \prod _{k=1}^i 2^{2^{h-k}} &= 2^{\sum _{k=1}^i 2^{h-k}} = 2^{2^h-2^{h-i}}\\ \overset{h}{\sum_{i=1} }\prod_{k=1}^i 2^{2^{h-k}}&=\overset{h}{\sum _{i=1} }2^{2^h-2^{h-i}}\\ \\ \\ j&=2^h-2^{h-i}\\ i&=1\Rightarrow j=2^h-2^{h-1}=\left(1-\frac{1}{2}\right) 2^h=\frac{2^h}{2}=2^{h-1}\\ i&=h\Rightarrow j=2^h-2^0=2^h-1\\ \sum_{i=1}^h 2^{2^h-2^{h-i}} &= \sum_{j=2^{h-1}}^{2^h-1} 2^j\\ &=\sum_{j=0}^{2^h-1} 2^j-\sum_{j=0}^{2^{h-1}-1} 2^j\\ &=\frac{2^{2^h}-1}{2-1}-\frac{2^{2^{h-1}}-1}{2-1}\\ &=2^{2^h}-2^{2^{h-1}}\\ &=u-\sqrt{u}\\ \\ \\ \overset{h}{\sum _{i=1} }\prod _{k=1}^i 2^{2^{h-k}}&=u-\sqrt{u} \end{align*} For simplicity I used variable $u={2^2}^h$.