# A number theory problem by Áron Bán-Szabó

\begin{aligned} {\color{#D61F06}{6}} & \mid (6-1)!=5\times 4\times 3\times 2\times 1 \\ {\color{#3D99F6}{7}} & \not\mid (7-1)!=6\times 5\times 4\times 3\times 2\times 1 \\ {\color{#D61F06}{8}} & \mid (8-1)!=7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{#D61F06}{9}} & \mid (9-1)!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{#D61F06}{10}} & \mid (10-1)!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\ {\color{#3D99F6}{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{#D61F06}{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{aligned}

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 $k$ ($>5$), ${\color{#D61F06}{k}}\mid (k-1)!=(k-1)\times (k-2)\times \dots \times 1$

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