A number theory problem by Áron Bán-Szabó

\[\begin{align} {\color{red}{6}} & \mid (6-1)!=5\times 4\times 3\times 2\times 1 \\ {\color{blue}{7}} & \not\mid (7-1)!=6\times 5\times 4\times 3\times 2\times 1 \\ {\color{red}{8}} & \mid (8-1)!=7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{red}{9}} & \mid (9-1)!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{red}{10}} & \mid (10-1)!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{blue}{11}} & \not\mid (11-1)!=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{red}{12}} & \mid (12-1)!=11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \end{align}\]

The red numbers are the composite numbers (bigger than \(5\)), the blue numbers are the prime numbers (bigger than \(5\)).

Is it true, that for any composite positive integer \(k\) (\(>5\)), \[{\color{red}{k}}\mid (k-1)!=(k-1)\times (k-2)\times \dots \times 1\]

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