On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 &  0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} &  1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} &  \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots &  \vdots & \ddots & \ddots & \vdots \\
\dfrac{1}{(n-1)!} &  \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\
\dfrac{1}{n!} & \dfrac{1}{(n-1)!} &  \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!}
\end{pmatrix}
=\dfrac{1}{n!}.
$$
 A: Hint. In general, let $d_0=d_1=1$ and let $(a_k)_{k=1,2,\ldots}$ be any sequence of numbers. For every $n\ge2$, denote by $d_n$ the determinant of the $n\times n$ Toeplitz-Hessenberg matrix
$$
\begin{pmatrix}
a_1 &1 &0 &0 &\cdots &0\\
a_2 &a_1 &1 &0 &\cdots &0\\
a_3 &a_2 &a_1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 &1\\
a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1
\end{pmatrix}.
$$
If one expands the determinant by the first column, one obtains
$$
d_n=-\sum_{k=1}^n(-1)^ka_kd_{n-k}.
$$
A: Hints
Prove it by induction.
At each step, expand by minors along the top row.
At the end, think about the binomial theorem.
A: HINT.-By property of determinants, we lower the order of n to (n-1)  as follows
$$\Delta_n=\det\begin{pmatrix}
1 &1 &0 &0 &\cdots &0\\
a_2 &1 &1 &0 &\cdots &0\\
a_3 &a_2 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 &1\\
a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
$$\Delta_n=\det\begin{pmatrix}
1 &0 &0 &0 &\cdots &0\\
a_2 &1-a_2 &1 &0 &\cdots &0\\
a_3 &a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_n &a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
$$\Delta_n=\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\
a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
On the other hand one has $\Delta_n=\dfrac{1}{n!}=\dfrac{1}{n}\dfrac{1}{(n-1)!}=\dfrac 1n\Delta_{n-1}$ so we have to prove that
$$\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\
a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}=\dfrac 1n\Delta_{n-1}$$ $$\dfrac 1n\Delta_{n-1}=\det\begin{pmatrix}\dfrac 1n &1 &0 &\cdots &0\\
\dfrac{1}{n}a_2 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
\dfrac 1na_{n-2} &a_{n-3} &\cdots &1 &1\\
\dfrac 1na_{n-1} &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
Note that the columns in these two last determinants  are all equal excepting the first.
Can you apply comfortably induction now to prove that both two last determinants are equal? 
