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I am having trouble proving that

$$\det \begin{pmatrix} \dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\ \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\ \dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\ \dfrac{1}{n!} & \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} \end{pmatrix} =\dfrac{1}{n!}. $$

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    $\begingroup$ Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. $\endgroup$ – saulspatz Jul 9 at 21:39
  • $\begingroup$ I'll reformulate my comment: Would you mind writting the next-to-last row? Is it: $\left(\frac{1}{(n-1)!}\cdots\frac{1}{1!} 1\right)$ or just $\left(\frac{1}{(n-1)!}\cdots\frac{1}{1!} 0\right)$? $\endgroup$ – miraunpajaro Jul 9 at 22:13
  • $\begingroup$ @miraunpajaro It is supposed to be the former. All the elements on the superdiagonal are $1$. $\endgroup$ – saulspatz Jul 9 at 22:24
  • $\begingroup$ @saulspatz Thanks! I was confused about the weird behaviour of the last row $\endgroup$ – miraunpajaro Jul 9 at 22:27
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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 $$ d_n=-\sum_{k=1}^n(-1)^ka_kd_{n-k}. $$

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Hints

Prove it by induction.

At each step, expand by minors along the top row.

At the end, think about the binomial theorem.

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  • $\begingroup$ I don't see how the expansion along the first row helps. Did you mean expansion along the bottom row or the first column? $\endgroup$ – user1551 Jul 10 at 0:15
  • $\begingroup$ I meant the first row, so at the first stage, there are two determinants, one of which is known by the induction hypothesis. $\endgroup$ – saulspatz Jul 10 at 0:18
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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?

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