<s>[INST] Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$ 
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq y$, $A_{n \times n}$.

The determinant expression appears to be Dickson polynomial of second kind.
Let $D_n$ be the determinant of $A_n$. We can see that the appropriate recurrence relation is $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Base cases:
$$D_1=x+y=\frac{x^2-y^2}{x-y}$$
$$
D_2=(x+y)^2-xy=x^2+xy+y^2=\frac{x^3-y^3}{x-y}
$$
Suppose that $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Then we need to prove that $$D_{n+1}=(x+y)D_{n}-xyD_{n-1}$$
Which can be developed as:
$$
D_{n+1}=(x+y)((x+y)D_{n-1}-xyD_{n-2})-xyD_{n-1}=
$$
$$
=(x+y)^2D_{n-1}-xy(x+y)D_{n-2}-xyD_{n-1}=
$$
$$
=(x^2+xy+y^2)D_{n-1}-xy(x+y)D_{n-2}=
$$
$$
=\frac{x^3-y^3}{x-y}D_{n-1}-xy(x+y)D_{n-2}
$$
I tried doing this up to $D_{n-6}$ in order to get any insights into possible simplification but I'm pretty stuck.
 [/INST] You did the base cases $n=1,2$, then you do the inductive step using strong induction; suppose your claim is true for any $m \leq n$, then using the recursive formula, for $n \geq 3$:
\begin{align}
D_{n+1} &=(x+y)D_n-xyD_{n-1} =\\ 
&=(x+y)\frac{x^{n+1}-y^{n+1}}{x-y} -xy\frac{x^n-y^n}{x-y} = \\
&= \frac{1}{x-y}(x^{n+2}-xy^{n+1}+yx^{n+1}-y^{n+2}-x^{n+1}y+xy^{n+1})= \\
&= \frac{x^{n+2}-y^{n+2}}{x-y}
\end{align}
which proves the claim for $m=n+1$. By induction, the claim is true for any $n$.
</s>