Let $x,y \in \mathbb{R}$.
Show that $$ x^{n+1}-y^{n+1}=(x-y)\sum_{k=0}^n x^k y^{n-k} $$
for all $n \in\mathbb{N}_0$
I need to prove this via induction.
My attempt: base case (k=0) = $$x^{0+1}-y^{0+1}=(x-y)(x^0 y^{n-0}) $$ $$x-y=(x-y)(y^n)$$ Here is where I get lost, does this disprove this? This statement is only true if $y^n = 1$? Doesn't this mean that this isn't true for all $x,y$ in $\mathbb R$?
On
The equation you're asking to prove is
$$x^{n+1}-y^{n+1}=(x-y)\sum_{k=0}^n x^k y^{n-k} \tag{1}\label{eq1}$$
for all $n \in \mathbb{N_{0}}$. As mentioned in multiple question comments and in Trevor Gunn's answer, the base case of $n = 0$ is true (as you use induction on the limit variable, not the summation one of $k$) since $x - y = (x - y)x^0 y^0$ and $x^0 = y^0 = 1$.
Continuing the induction proof, assume \eqref{eq1} is true for $n = m$ for some $m \ge 0$, i.e.,
$$x^{m+1}-y^{m+1}=(x-y)\sum_{k=0}^m x^k y^{m-k} \tag{2}\label{eq2}$$
The RHS of \eqref{eq1} for $n = m + 1$ is
$$\begin{equation}\begin{aligned} (x-y)\sum_{k=0}^{m+1} x^k y^{m+1-k} & = (x-y)\left(\sum_{k=0}^{m} x^k y^{m+1-k} + x^{m+1}y^{0}\right) \\ & = (x-y)\left(y\sum_{k=0}^{m} x^k y^{m-k} + x^{m+1}\right) \\ & = y\left((x-y)\sum_{k=0}^{m} x^k y^{m-k}\right) + (x-y)x^{m+1} \\ & = y\left(x^{m+1}-y^{m+1}\right) + x^{m+2} - yx^{m+1} \\ & = yx^{m+1} - y^{m+2} + x^{m+2} - yx^{m+1} \\ & = x^{m+2} - y^{m+2} \end{aligned}\end{equation}\tag{3}\label{eq3}$$
This is the LHS of \eqref{eq1} for $n = m + 1$. This shows \eqref{eq1} is true for $n = m + 1$ if it's true for $n = m$, and since it's true for $n = 0$, induction shows \eqref{eq1} is true for all $n \in \mathbb{N_{0}}$.
When you apply the principle of induction to an identity involving a sum, the base case and induction step is always in terms of the variable on the top of the summation ($n$) and not the indexing variable ($k$). So the base case is $n = 0$ which looks like
$$x^{0 + 1} - y^{0 + 1} = (x - y)\sum_{k = 0}^0x^ky^{0-k} = (x - y)(x^0y^0). $$