Arithmetic sequences have a common difference, where you add a constant to each term to get the next. Geometric sequences have a common ratio, where you multiply a constant to each term to get the next. Each one of these sort of sequences have a summation formula where you can sum the first $n$ terms without generating them.
For an arithmetic sequence $(a_i)_{i=1}^n$, we have
$$\begin{align} a+(a+d)+(a+2d)+\cdots+(a+(n-1)d) &= na+(1+2+\cdots+(n-1))d \\ &=na+\frac{(n-1)(n)}{2}d. \end{align}$$
For a geometric sequence $(a_i)_{i=1}^n$, we have
$$\begin{align} a+ra+r^2a+\cdots+r^{n-1}a &= s \\ ra+r^2a+r^3a+\cdots+r^{n}a &= rs\\ \phantom{d}&\\ \mbox{subtracting line 2 from line 1 gives us:}&\\ \\ r^na -a&= rs-s\\ (r^n-1)a &= (r-1)s\\ \\ \mbox{so, we get:}\\ \\ s &= \frac{r^n-1}{r-1}a \end{align}$$
What if we take this to the next natural step. Instead of adding a constant to get the next term, as in an arithmetic sequence, or multiplying a constant to get the next term, as in a geometric sequence, you instead raise each term to a constant power to get the next term. So, you would have $$a,a^e,(a^e)^e,((a^e)^e)^e,\dots$$ or, more elegantly, $$a,a^e,a^{e^2},a^{e^3},\dots,a^{e^n}$$ for some constant $e$. I feel like this would be called an exponential sequence, but that seems to be another name for a geometric sequence from google searches.
I was playing with trying to make a summation formula for the first $n$ terms of such a sequence, and quickly realized I was stuck. I am sure others have played with such a sequence before, but I couldn't find anything online for similar sequences. Does anyone have any ideas for how to proceed with generating a summation formula? Note that I am not asking what the formula is - I want to derive it myself (this is just for fun), but I cannot find a good starting point.
Thanks for any thoughts you all may have!
-Andrew
Edit: A user named Doug supplied a useful starting point on a comment that has since been deleted, suggesting an approach using partial sums. I wasn't able to look at it until late last night, and I spent a few hours working on it before I had to go to bed. Unfortunately, I got stuck in the weeds of the calculus on their approach, and since it is now gone I can't continue working on it. But, the starting point is sound I think.
Note that I will be using $c$ as the constant power, instead of $e$, because as users have pointed out, $e$ was a bad choice (Euler's constant).
If we think of the sum as a finite power series $S_n(x) = \sum_{i=1}^n x^{c^{i-1}}$, the question becomes finding a closed form of that finite power series. We notice that
$$S_n(x^c)-S_n(x)=x^{c^n}-x$$
This gives us a starting point similar to that in finding the summation formula of a geometric series. Unfortunately, the left-hand side doesn't lend itself well to being written as a multiple of $S_n(x)$. I have tried some tricks, seeing if I could take derivatives and end up with a solvable ODE, but I haven't had much luck.
I am going to keep thinking about this, but if anyone browsing the forum has any ideas, please feel free to share!
