Suppose

$\displaystyle x^n-y^n\ =\ (x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})\ \ldots\ \fbox1$

is true. Multiplying through by $\displaystyle x$ gives

$\displaystyle x^{n+1}-xy^n\ =\ (x-y)(x^n+x^{n-1}y+\cdots+x^2y^{n-2}+xy^{n-1})\ \ldots\ \fbox2$

while multiplying $\displaystyle \fbox1$ by $\displaystyle y$ gives

$\displaystyle x^ny-y^{n+1}\ =\ (x-y)(x^{n-1}y+x^{n-1}y^2+\cdots+xy^{n-1}+y^n)\ \ldots\ \fbox3$

$\displaystyle \fbox2+\fbox3$ gives

$\displaystyle x^{n+1}+y^{n+1}+x^ny-xy^n\ =\ (x-y)(x^n+2\left[x^{n-1}y+x^{n-2}y^2+\cdots+x^2y^{n-2}+xy^{n-1}\right]+y^n)$

But

$\displaystyle x^ny-xy^n$

$\displaystyle =\ xy(x^{n-1}-y^{n-1})$

$\displaystyle =\ xy(x-y)(x^{n-2}+x^{n-3}y+\cdots+xy^{n-3}+y^{n-2})$

$\displaystyle =\ (x-y)(x^{n-1}y+x^{n-2}y^2+\cdots+x^2y^{n-2}+xy^{n-1})$

Hence the extra terms on both sides should cancel, giving the result we need.

**Important note:** Because we assumed the result to be true for both $\displaystyle n$ and $\displaystyle n-1$ in order to prove it true for $\displaystyle n+1,$ the whole proof by induction should be completed by showing that the result is true for *both* $\displaystyle n=1$ *and* $\displaystyle n=2.$ (Showing that it is true for just $\displaystyle n=1$ is not good enough.)