Here is the expression:
$x^n-y^n$
According to my professor, this is factorable, and an example for telescoping in mathematics. I've gleaned some sense of what Telescoping is from this answer: Mathematical Telescoping. Could someone walk me through how the above expression can be factored, and how it exhibits telescoping (and why telescoping is salient to begin with?).
The way telescoping works is really simple. As an example I will show you how one can derive the formula to calculate the sum of the first $n$ terms of a geometric series.
We want to find $S_n = 1 + x + x^2 + x^3 + \cdots + x^n$
We first compute $xS_n = x + x^2 + x^3 + \cdots + x^n + x^{n+1}$
And now we calculate $xS_n - S_n$:
$$\begin{align*} xS_n - S_n =\ &\color{red}{x + x^2 + x^3 + \cdots + x^n} + x^{n+1}\\ - (1 +&\color{red}{ x + x^2 + x^3 + \cdots + x^n}) \end{align*} \iff\\ S_n(x - 1) = - 1 - x^{n+1}\\ S_n = \frac{-1-x^{n+1}}{x - 1} = \frac{1 + x^{n+1}}{1 - x}$$
The telescoping bit was when the factors in red canceled out.
We can apply a similar technique for $(x^n - y^n)$. Motivated by the fact that $(x^2 - y^2) = (x - y)(x + y)$ we may try to write $(x^n - y^n)$ as $(x - y)(\text{some factors here})$. The easiest factors to guess are the $x^{n-1}$ and $y^{n-1}$ that will not be cancelled out. Let us call $P$ the part that gets factored out to which we will apply the telescoping method. We already now $P = x^{n-1} + a_2 + \cdots + a_k + y^{n-1}$ and our aim is to have $(x - y)P = (x^n - y^n)$,
$$\begin{align*} (x - y)P = xP - yP =\\ x(x^{n-1} + a_2 + a_3 + \cdots + a_k &+ y^{n-1})\\ - (y(x^{n-1} + a_2 + a_3 + \cdots &+ a_k + y^{n-1}))\iff\\ (x^{n} + xa_2 + xa_3 + \cdots + xa_k &+ xy^{n-1})\\ - (yx^{n-1} + ya_2 + ya_3 + \cdots &+ ya_k + y^{n}) \end{align*}$$
And you get that
$$\begin{cases} xa_2 = yx^{n-1}\\ ya_i = xa_{i+1}\\ ya_k = y^{n-1}x \end{cases}$$
Using the second equation, you can write $a_{i+1} = \frac{y}{x}a_i$ and using the first case you can get that $a_2 = yx^{n-2}$ and then we can get, by the third case, $a_k = y^{n-2}x$. Each iteration your exponent on $y$ increases by 1 and your $x$ exponent decreases by 1 so we can see that $k = n-2$ and rewrite $a_i = y^{i-1}x^{n-i}$ like that. Thus we have
$$P = \sum_{i=1}^{n} a_i = \sum_{i=1}^{n} y^{i-1}x^{n-i}$$
and we should note that this, for $i = 1$ and $i = n$ goes with our initial guesses. One can now show that $(x - y)P = (x^n - y^n)$:
$$(x - y)P = (x - y)\sum_{i=1}^{n} y^{i-1}x^{n-i} = x\sum_{i=1}^{n} y^{i-1}x^{n-i} - y\sum_{i=1}^{n} y^{i-1}x^{n-i} =\\ = \sum_{i=1}^{n} y^{i-1}x^{n-i+1} - \sum_{i=1}^{n} y^{i}x^{n-i}$$
Now with a careful manipulation of the first summation's limits we can use the telescoping method! We rewrite
$$\sum_{i=1}^{n} y^{i-1}x^{n-i+1} = \sum_{i=1}^{n} y^{i-1}x^{n-(i-1)} = \sum_{i=0}^{n-1} y^{i}x^{n-i}$$
And putting everything together yields
$$(x-y)P = \sum_{i=0}^{n-1} y^{i}x^{n-i} - \sum_{i=1}^{n} y^{i}x^{n-i} = x^n - y^n$$ because all terms cancel one-to-one except the first of the first summation and the last of the last summation.
In my view this is the result of a telescoping sum. From the formula of the partial sum of a geometric series it can be derived that
$(x-y)\cdot \sum_{k=0}^{n-1} x^k\cdot y^{n-k-1}=x^n-y^n$
$\sum_{k=0}^{n-1} \left( x^{k+1}\cdot y^{n-k-1}-x^k\cdot y^{n-k}\right)$
$=x\cdot y^{n-1}-y^{n}+x^2\cdot y^{n-2}-x\cdot y^{n-1}+x^3\cdot y^{n-3} -x^2\cdot y^{n-2}+\ldots+x^{n-1}\cdot y -x^{n-2}\cdot y^2 +x^n-x^{n-1}\cdot y $
I think you see that only $x^n-y^n$ remains. You can see that the most positive summands have a corresponding negative summand as well. The consequence is that a series can be simplified by using the fact that the sum of these pairs of summands are $0$.
