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If we let $z=xy$, we have: $$z=\sum_{i=1}^{y}x$$ So multiplication can be written as iterative sum. Likewise, if we have $z=x^y$, then we can write: $$z=\prod_{i=1}^{y}x$$ But how to write the exponent in sum form instead of product form?

I tried to experiment by writing for example $5^2=(5+5+5+5+5), 2^4=((2+2)+(2+2))+((2+2)+(2+2))$, but I came up with nothing.

Edit: From the conversation in the comments, you can write exponent as nested summations:

$$z=\sum_{i=1}^{x}\sum_{j=1}^{x}\sum_{k=1}^{x}\dots\text{(y times)}\dots x$$

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  • $\begingroup$ Would you be satisfied with a recursive formula? Otherwise some potentially involved sums would comprise a general form. $\endgroup$
    – Xoque55
    Commented Feb 20, 2017 at 2:57
  • $\begingroup$ Whatever uses sums only, I'd be happy with. $\endgroup$
    – KKZiomek
    Commented Feb 20, 2017 at 2:58
  • $\begingroup$ Applying both your observations at the same time, consider $z=x^y = x^1 x^{y-1} = \sum\limits_{i=1}^{x^{y-1}} x$ which could be swapped around if you don't like the "large" upper index, it could also be $\sum\limits_{i=1}^{x} x^{y-1}$ $\endgroup$
    – Xoque55
    Commented Feb 20, 2017 at 3:02
  • $\begingroup$ The summation itself uses exponentation, so it isn't the best that can be done. $\endgroup$
    – KKZiomek
    Commented Feb 20, 2017 at 3:03
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    $\begingroup$ What about nested sums? Such as if $z = x^y$, then you could iterate until you've hit a desired number of inner summations: $z = \sum\limits_{i = 1}^{x} x^{y-1} = \sum\limits_{i = 1}^{x} \left( \sum\limits_{j = 1}^{x} x^{y-2} \right) = \sum\limits_{i = 1}^{x} \left( \sum\limits_{j = 1}^{x} \left( \sum\limits_{k = 1}^{x} x^{y-3} \right) \right)$ $\endgroup$
    – Xoque55
    Commented Feb 20, 2017 at 3:10

1 Answer 1

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Does this work?:

Let $z = x^y$. This will work for $x,y \in \mathbb{N}$.

$$ z = x^y \\ = \sum\limits_{i_1 = 1}^{x} x^{y-1} \\= \sum\limits_{i_1 = 1}^{x} \left( \sum\limits_{i_2 = 1}^{x} x^{y-2} \right) \\ = \sum\limits_{i_1 = 1}^{x} \left( \sum\limits_{i_2 = 1}^{x} \left( \sum\limits_{i_3 = 1}^{x} x^{y-3} \right) \right) \\ = \underbrace{\sum\limits_{i_1 = 1}^{x} \left( \sum\limits_{i_2 = 1}^{x} \left( \sum\limits_{i_3 = 1}^{x} \left( \cdots \left( \sum\limits_{i_y = 1}^{x} 1 \right) \right) \right) \right)}_{y \text{ times}} $$

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  • $\begingroup$ Well, I realized that no one in the history ever thought of a compact notation that tells how many times something is nested, so basically your answer is possibly the answer. $\endgroup$
    – KKZiomek
    Commented Feb 20, 2017 at 3:26
  • $\begingroup$ So now you see the power of the notation for exponents! [pun intended] $\endgroup$
    – Xoque55
    Commented Feb 20, 2017 at 4:08

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