Proving that $u_{n}\in\mathbb{Z}\forall n\ge 0$, such that $\det \begin{bmatrix}u_{n} & u_{n+1} \\ u_{n+2}&u_{n+3}\end{bmatrix}=n!\forall n\ge 0$

The pattern is $$ u_2 = 1 = 1 u_0 \\ u_3 = 2 = 2 u_1 \\ u_4 = 3 = 3 u_2 \\ u_5 = 8 = 4 u_3 \\ u_6 = 15 = 5 u_4 \\ u_7 = 48 = 6 u_5 $$ and generally, $u_n = (n-1) u_{n-2}$ for $n \ge 2$. It follows that $$ u_n = (n-1)!! = (n-1)(n-3)(n-5) \cdots $$ for $n \ge 1$ where $!!$ denotes the “double factorial.” It is now easy to verify that $$ \det\begin{bmatrix}u_n & u_{n+1} \\ u_{n+2} & u_{n+3}\end{bmatrix} = u_n u_{n+1}\det\begin{bmatrix}1 & 1 \\ {n+1} & {n+2}\end{bmatrix} = u_n u_{n+1} = n! \, . $$ Apart from the initial term $u_0 = 1$ this is A006882 in the On-Line Encyclopedia of Integer Sequences®.