Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the following lines:
Since $\gcd(6,5) = 1$ this means that if we multiply $5$ with all the numbers from $1$ to $5$ we have to get all the numbers in different order.
I.e.
The numbers less than $6$: $1,2,3,4,5$.
Multiply the above by $5$: $5,4,3,2,1$ i.e. we get the same set of numbers but in different order.
Since: $1\cdot 2\cdot 3\cdot 4\cdot 5 = 5\cdot 4\cdot 3\cdot 2\cdot 1$ we also have:
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow  1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$
Which is actually wrong.
What is the problem in my approach and thought process?
 A: The fallacy in your argument is in the last step, where you "cancel" $1\cdot2\cdot3 \cdot 4\cdot 5$ from each side of the congruence.
As in ordinary arithmetic, you can only cancel factors that aren't $0$. (In fact, you can only reliably cancel factors that are relatively prime to the modulus.)
So in this case the assertion you are trying to prove is correct but your argument is not. The argument does work for arithmetic with a prime modulus.
A: A most immediate thing to note is that $(m-1)\equiv -1 \pmod m$ always.
So $(m-1)^{m-1} \equiv (-1)^{m-1}\pmod m$.  And $(-1)^{m-1} = -1$ if $m-1$ is odd and $(-1)^{m-1} = 1$ if $m-1$ is even.  And as $-1 \equiv m-1\pmod m$ and $1\equiv m+ 1\pmod m$ we realize what you probably heard was
$(m-1)^{m-1} \equiv (-1)^{m-1} \equiv \begin {cases}1\equiv m+1 \pmod m& m\text{ is odd}\\-1\equiv m-1 \pmod m& m\text{ is even}\end{cases}$

"The numbers less than 6: 1,2,3,4,5.
Multiply the above by 5: 5,4,3,2,1 i.e. we get the same set of numbers but in different order."

Note:  That is because $5\equiv -1$ so $1,2,3,4,5$ times $5$ will be $-1,-2,-3,-4,-5$ which are $5,4,3,2,1$.

$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow  1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$ Which is actually wrong.

But $1\cdot 2\cdot 3\cdot 4\cdot 5\not \equiv 1$.
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 0$.
And you od have $1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow  1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 0 \equiv 5^5\cdot 0$
Ah.... I see the issue...

Wilson's theorem: $(p-1)! \equiv -1 \pmod p$ !!!IF $p$ IS PRIME!!!

But if $m$ is composite that the theorem does not hold  $(m-1)! \not \equiv 1 \pmod m$.
In fact:

If $m> 4$ is composite then $(m-1)! \equiv 0 \pmod m$.

Pf:  If $m$ is composite then $m = dk$ for some $d,k$ which are neither $1$ nor $m$ and so are components of $(m-1)!$. So $m|(m-1)!$ if $m$ is composite and $(m-1)!\equiv 0 \pmod m$.
....unless!..... $m = p^2$ for some prime.  Then $p$ only occurs as a component once.  But then $p < 2p < .....< p^2$ and $p$ and $2p$ are components.  In which case $p\cdot 2p = 2p^2 = 2m|(m-1)!$ and so $m|(m-1)!$ and $(m-1)! \equiv 0 \pmod m$.
....unless! .... $p=2$ and $m =2^2 = 4$.  But in that case $(4-1)! = 6$ and $6\equiv 2\pmod 4$.  That is the exception.  But if $m > 4$ there is no exception.
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Oh.... You weren't doing Wilsons theorem... which would have been negative anyway....
You were assuming $ac \equiv bc \pmod m \implies a \equiv b\pmod m$.
No.  Division is one of the operations you can't do via modular arithmetic (unless $m$ is prime).
If $ac \equiv bc \pmod m$ then $m|ac -bc=(a-b)c$.  But that does not mean $m|a-b$ unless $\gcd(m,c) =1$.  For example if we let $a= 5$ and $b=3$ and $c= 3$ and $m = 6$ we have $15 \equiv 9\pmod 6$ and $6|15-9 = (5-3)3 = 2\times 3$ but we do not have $6|2$.  So we do not have $5\equiv 3 \pmod 6$.
So we can't do division.  (Unless $m$ is prime)
A: your statement $$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow  1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$$
is not corrrect .
Note that $$1.2.3.4.5\equiv 0 \pmod 6 $$
