On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 &  0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} &  1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} &  \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots &  \vdots & \ddots & \ddots & \vdots \\
\dfrac{1}{(n-1)!} &  \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\
\dfrac{1}{n!} & \dfrac{1}{(n-1)!} &  \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!}
\end{pmatrix}
=\dfrac{1}{n!}.
$$

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? 
