Rewriting exponents as sigmas, is this a thing? Is this a thing? If not, can anyone help me out on this?
So I saw this a while back.$$\sum_{i=0}^n(2n + 1)  = n^2$$ For positve n.
This is interesting, and I wondered if I could write any n^x in sigma notation.
Here was my thought process. Take the difference of the powers, to find a pattern.
This for $n^2$ gives:
0 1 4 9 16 25...
1 3 5 7 9...
2 2 2 2...
So this says that the difference of consecutive squares are odd numbers that increase by two. $2^7$ could be said as $1+3+5+7+9+11+13$.
For cubes I did the same.
0 1 8 27 64 125...
1 7 19 37 61...
6 12 18 24...
6 6 6...
This says that that the difference of the differences of the differences(pls no more redundancy) of a set of cubes is 6. $6^3$ can be said as $1+7+19+37+61+91$.
This in sigma notation is $$\sum_{i=0}^n(2n + 1)  = n^2$$
and$$\sum_{i=0}^n((\sum_{j=0}^{i}6j) + 1)  = n^3$$
I programmed a tool to help me find the num that the differences boil down to.
We know that 2's is 2, 3's is 6.
I calculated it for 3 more then found something.
4: 24
5: 120
6: 720
These were all factorials. $5! = 120, 6! =720$ and so on.
These equations will all contain these numbers.
I think that n to the xth power would have x sigmas that sum each other, with the last sigma summing x!. This is most likely close but wrong.
I'm missing a lot of info on this, I need to think more.

Here are a few ideas:

$$(n+1)^2-n^2~=~2n+1,\qquad\qquad(n+1)^3-n^3~=~3n^2+3n+1,\qquad\text{etc.}$$

$$\sum_{n=0}^Nn~=~\frac{N(N+1)}{2!}~,\qquad\qquad\sum_{k=0}^N~\sum_{n=0}^kn~=~\frac{N(N+1)(N+2)}{3!}$$
$$~$$
$$\sum_{k=0}^N~\sum_{m=0}^k~\sum_{n=0}^mn~=~\frac{N(N+1)(N+2)(N+3)}{4!}~,\qquad\quad\text{etc.}$$

$$\sum_{n=0}^Nn~=~\frac{N(N+1)}2~,\qquad\qquad\sum_{n=0}^Nn(n+1)~=~\frac{N(N+1)(N+2)}3$$
$$~$$
$$\sum_{n=0}^Nn(n+1)(n+2)~=~\frac{N(N+1)(N+2)(N+3)}4~,\qquad\quad\text{etc.}$$
