Wilson's Theorem/Examples/10 does not divide (n-1)!+1

Example of Use of Wilson's Theorem

For all $n \in \Z_{>0}$, $10$ is not a divisor of $\paren {n - 1}! + 1$.

For the first few $n$ we see:

$\, \ds 10 \nmid \, $

$\ds 2$

$=$

$\ds \paren {1 - 1}! + 1$

where $\nmid$ denotes non-divisibility

$\ds $

$=$

$\ds \paren {2 - 1}! + 1$

$\, \ds 10 \nmid \, $

$\ds 3$

$=$

$\ds \paren {3 - 1}! + 1$

$\, \ds 10 \nmid \, $

$\ds 7$

$=$

$\ds \paren {4 - 1}! + 1$

$\, \ds 10 \nmid \, $

$\ds 25$

$=$

$\ds \paren {5 - 1}! + 1$

Now consider $n > 5$.

We have that:

$\ds 2$

$\divides$

$\ds \paren {n - 1}!$

where $\divides$ denotes divisibility

$\ds 5$

$\divides$

$\ds \paren {n - 1}!$

$\ds \leadsto \ \ $

$\ds 10$

$\divides$

$\ds \paren {n - 1}!$

as $10 = \lcm \set {2, 5}$

Hence $10 \nmid \paren {n - 1}! + 1$.

$\blacksquare$

1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-3}$ Wilson's Theorem: Exercise $2$