Does this pattern have anything to do with derivatives? In 6th grade I was first introduced to the idea of a function in the form of tables. The input would be "n" and the output "$f_n$" would be some modification of the input. I remember finding a pattern in the function "f(n)=n^2". Here is what the table looked like:
\begin{array}{|c|c|}
\hline
n& f_n\\ \hline
1&1 \\ \hline
2&4\\ \hline
3&9\\ \hline
4&16\\ \hline
5&25\\ \hline
...&...\\ \hline
n&n^2\\ \hline
\end{array}
I would then take the outputs $f_n$ and find the differences between each one: $f_n-f_{n-1}$. This would produce:
\begin{array}{|c|c|}
\hline
n& f(n)-f(n-1)\\ \hline
1&1 \\ \hline
2&3\\ \hline
3&5\\ \hline
4&7\\ \hline
5&9\\ \hline
...&...\\ \hline
\end{array}
Repeating this process (of finding the differences) for the outputs of  $f_n-f_{n-1}$ would yield a continuous string of $2$s. As a 6th grader I called this process 'breaking down the function' and at the time it was just another pattern I had found.
Looking back at my work as a freshman in high school, I realize that the end result of 'breaking down a function' corresponds to the penultimate derivative (before the derivative equals zero). For example: breaking down $y=x^3$ gives a continuous string of $6$s, and the third derivative of $x^3$ is 6 (while the 2nd derivative is 6x). 
Is there any significance to this pattern found by finding the differences between each output of a function over-and-over again? Does it have anything to do with derivatives? I know my question is naive, but I'm only a high school freshman in algebra II. A non-calculus (or intuitively explained calculus concepts) answer would be very helpful [note that I used an online derivative calculator to find the derivatives of these functions and I apologize for any incorrect calculus terminology].

Yes it is related to finite calculus, but others have answered why and how while no one mentioned it is also related to sum of arithmetic sequences so I will.
Let's have arithmetic sequence where $a_1 = 1$ and $d = 2$ (so basically it's sequence of odd numbers).
Sum of first $n$ elements of that sequence (denoted as $S_n$) is:
$$S_n = \frac{2a_1 + (n - 1)d}{2}n = \frac{2 + 2(n - 1)}{2}n = (1 + n -1)n = n^2$$
It is obvious that $S_n - S_{n-1} = a_n$ and because of $S_n=n^2$ we have:
$$n^2-(n-1)^2=S_n-S_{n-1}=a_n$$
In your table for $n$ you got that $n^2-(n-1)^2$ is nth odd number which is the same as formula which I got.
