Calculate the determinant of given matrix The matrix $A_n\in\mathbb{R}^{n\times n}$ is given by
$$\left[a_{i,j}\right] = \left\lbrace\begin{array}{cc}
1 & i=j \\
-j & i = j+1\\
i & i = j-1 \\
0 & \text{other cases}
\end{array} \right.$$
I already showed that it holds
$$\det{A_n}= \det{A_{n-1}}+\left(n-1\right)^2\cdot\det{A_{n-2}}$$
However, can we find an explicit expression for the determinant of $A_n$?
 A: I had an idea by myself right now...
It holds


*

*$\det(A_1) = 1$

*$\det(A_2) = 2$

*$\det(A_3) = 6$

*$\det(A_4) = 24$

*$\dots$


Thus I assume the determinant of $A_n$ is given by $n!$

Proof: (by induction over $n$)


*

*The statement is true for $n=1\dots 4$

*Now let it be true for $n-1$ and $n-2$ then:


$$\begin{array}{rcl}
\det(A_n) &=& \det(A_{n-1}) + \left(n-1\right)^2\det(A_{n-2})\\
&=& (n-1)! + (n-1)^2\cdot(n-2)! \\
&=& (n-1)! + (n-1)\cdot(n-1)! \\
&=& (1+(n-1))\cdot(n-1)! \\
&=& n!
\end{array}$$
Hence $\det(A_n) = n!$ is true for every $n\geq 1$
A: Put $a_n=\det A_n$ and 
$$ f(x)=\sum_{n=1}^{+\infty}a_n \frac{x^n}{n!},\tag{1}$$
in order that:
$$f'(x) = \sum_{n=0}^{+\infty}a_{n+1}\frac{x^{n}}{n!},\tag{2}$$
$$x\,f(x) = \sum_{n=2}^{+\infty}n\, a_{n-1} \frac{x^{n}}{n!},$$
$$f(x)+x\,f'(x) = \sum_{n=2}^{+\infty}n^2\, a_{n-1} \frac{x^{n-1}}{n!},$$
$$x\,f(x)+x^2\,f'(x) = \sum_{n=2}^{+\infty}n^2\, a_{n-1} \frac{x^{n}}{n!}.\tag{3}$$
Now $a_{n+1}=a_n+n^2\,a_{n-1}$ gives:
$$f'(x) = f(x) + x\, f(x) + x^2\,f'(x),$$
or:
$$ (1+x)\,f(x) = (1-x^2)\, f'(x)$$
$$ f(x) = (1-x)\, f'(x)\tag{4}. $$
The solutions of this differential equation are $f(x)=\frac{K}{1-x}$,
from which we get that $a_n = K\,n!$. Since $a_1=1$, $a_n=n!$.
