How do I finish this induction proof in linear algebra? So I'm given an n*n matrix for  $a_0$
$$
A_{n}\left(a_{0}, \ldots, a_{n-1}\right)=\left(\begin{array}{ccccc}
{t} & {0} & {\cdots} & {0} & {a_{0}} \\
{-1} & {t} & {\ddots} & {\vdots} & {a_{1}} \\
{0} & {-1} & {\ddots} & {0} & {\vdots} \\
{\vdots} & {\ddots} & {\ddots} & {t} & {a_{n-2}} \\
{0} & {\ldots} & {0} & {-1} & {t+a_{n-1}}
\end{array}\right)
$$
for $n \in \mathbb N$. Then we define  $d_{n}\left(a_{0}, \ldots, a_{n-1}\right)=\operatorname{det}\left(A_{n}\left(a_{0}, \ldots, a_{n-1}\right)\right)$. So far, I have already shown that  $d_{n}\left(a_{0}, \ldots, a_{n-1}\right)=t d_{n-1}\left(a_{1}, \ldots, a_{n-1}\right)+a_{0}$, so now what remains to be shown is using induction. 
The question asks to show inductively that $d_{n}\left(a_{0}, \ldots, a_{n-1}\right)=t^{n}+a_{n-1}t^{n-1} + \dots + a_1 t + a_0$. I am quite confused about how to do this but I have tried it in the following way: 
Since we know the recursive formula for $d_n(a_0, \dots , a_{n-1})=td_{n-1} (a_1, \dots , a_{n-1}) +a_0$, it follows that $d_{n+1}(a_0,\dots, a_n) = td_n(a_1 \dots , a_n) + a_0$. 
So then $td_{n+1}(a_1 , \dots , a_n) + a_0 = t^{n+1} + a_n t^n + \dots + a_1 t + a_0$ which if you pull the $t$ out and use the inductive hypothesis on then it leaves $td_{n+1}(a_1 , \dots , a_n) = t(t^n + a_n t^{n-1} + a_{n-1}t^{n-1} + \dots + a_2 t + a_1)$. This would be fine to use the hypothesis on and those solve the problem, but the indexes on my a's are shifted up by one. How do I resolve this? 
 A: You have all the piece, you just mixed up the indices.
Lets start at the beginning, is it true for $n = 2$
$$A_2(a_0,a_1) = \begin{pmatrix}t&a_0\\-1&t+a_1\end{pmatrix}$$
$$d_2(a_0, a_1) = t^2+a_1 t + a_0$$
Suppose now it is true for $n$, $d_n(a_0,\ldots,a_{n-1}) = t^t+a_{n-1}t^{n-1}+\cdots+a_1 t + a_0$, we will show it is also true for $n+1$
$$
A_{n+1}\left(a_{0}, \ldots, a_{n-1},a_n\right)=\left(\begin{array}{ccccc}
{t} & {0} & {\cdots} & {0} & {a_{0}} \\
{-1} & {t} & {\ddots} & {\vdots} & {a_{1}} \\
{0} & {-1} & {\ddots} & {0} & {\vdots} \\
{\vdots} & {\ddots} & {\ddots} & {t} & {a_{n-1}} \\
{0} & {\ldots} & {0} & {-1} & {t+a_{n}}
\end{array}\right)
$$
By developping the determinant by the first line, we have
$$d_{n+1}\left(a_{0}, \ldots, a_{n-1},a_n\right) = td_n\left(a_{1}, \ldots, a_{n-1},a_n\right)+a_0$$
We suppose it is true for $n$, so we develop $d_n$
$$d_{n+1}\left(a_{0}, \ldots, a_{n-1},a_n\right) = t\left(t^n+a_{n}t^{n-1} + \cdots + a_{2}t+a_1\right)+a_0$$
Distribute $t$ inside the parenthesis give the desired result
$$d_{n+1}\left(a_{0}, \ldots, a_{n-1},a_n\right) = t^{n+1}+a_{n}t^{n} + \cdots + a_{2}t^2+a_1t+a_0$$
Hope it helps.
