Is there anything special about this summation I found that calculates out the square, cube, fourth, etc. power of any integer? let me start by saying that my formatting may be way off, but it's the best I can do, and has little to do with the question, and I will make sure I am as clear as humanly possible, including showing all my work 
So I was playing with squares and stuff, I found that I could calculate out every successive perfect square with: $$\sum_{n=0}^{x-1} (2n+1) = x^2$$ where $$ 0 \le n\le(x-1)|_{||}| 1\le(n+1)\le x$$ 
For example: $5^2=\sum_{n=0}^4(2n+1)=1+3+5+7+9=25$
and then found through trial and error that I could also calculate out successive cubes with: $$ x^3=\sum_{n=0}^{x-1}(3n^2+3n+1) $$ with the same rules for n.
For example: $5^3=\sum_{n=0}^4 (3n^2+3n+1)=1+7+19+37+61=125$
Then I discovered that the cubic formula can be factored so that: $$x^3=\biggr(\sum_{n=0}^{x-1}\bigr((2n+1)*(n+1)\big)\biggr) $$
I noticed each even power can be represented by: $x^{m_{even}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac m 2}$ 
While odd powers can be represented with: $x^{m_{odd}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac{m-1} 2}*(n+1)$ 
where $(n+1)=x$ , but only the final term is replaced with $x$. 
Which eventually lead me to an interesting relation (at least to me): $$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg)$$
I get that I'm essentially just multiplying squares and cubes together and essentially saying that: $$x^m=\big((x^2)^{\frac {m-3} 2}\big)*x^3$$ and all the math works out to give you: $$x^{m-3}=x^{m-3}$$
However, I want to know if this is something unique, or are there many other ways to represent $x^m$ than what I've found? Is there anything useful somebody could do with something like this, or is it only good for looking cool? 
p.s. Sorry if what I tried to convey didn't make any sense, I'm not familiar with how I'm supposed to phrase things in math.

You are implicitly using telescoping series with your sums giving the integer squares and cubes. For the first case, you have
$$\begin{equation}\begin{aligned}
& \sum_{n=0}^{x-1}(2n+1) \\
& = \sum_{n=0}^{x-1}((n^2 + 2n+1) - n^2) \\
& = \sum_{n=0}^{x-1}((n+1)^2 - n^2) \\
& = (1^2 - 0^2) + (2^2 - 1^2) + \ldots + ((x-1)^2 - (x-2)^2) + (x^2 - (x-1)^2) \\
& = x^2 - 0^2 \\
& = x^2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The final result occurs because all of the terms cancel each other out, except for the $x^2$ and $0$ terms. Similarly, you also have
$$\begin{equation}\begin{aligned}
& \sum_{n=0}^{x-1}(3n^2+3n+1) \\
& = \sum_{n=0}^{x-1}((n^3 + 3n^2+3n+1) - n^3) \\
& = \sum_{n=0}^{x-1}((n+1)^3 - n^3) \\
& = (1^3 - 0^3) + (2^3 - 1^3) + \ldots + ((x-1)^3 - (x-2)^3) + (x^3 - (x-1)^3) \\
& = x^3 - 0^3 \\
& = x^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Regarding your statement of
$$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg) \tag{3}\label{eq3A}$$
It's true, as you determined by substituting and simplifying. As for any particular use for this, I don't know of any offhand. Also, regarding other ways to express $x^m$, there are many possibilities, with the particular choice often depending on what you're checking on or trying to determine.
