Remainder of $35! $ when divided by $41$ I use $p=41 $ is a prime, so 

$\begin{align}
(41-1)! &amp;\equiv -1 \mod 41\\
(40)! &amp;\equiv -1 \mod 41\\
40\times39\times38\times37\times36\times(35)! &amp;\equiv -1 \mod 41\\
(-1)\times(-2)\times(-3)\times(-4)\times(-5)\times(35)! &amp;\equiv -1 \mod 41\\
120(35)! &amp;\equiv +1 \mod 41\\
(-3)(35)! &amp;\equiv +1 \mod 41\\
(3)(35)! &amp;\equiv  -1 \mod 41 \\
(3)(35)! &amp;\equiv  -42 \mod 41 \ \to \div 3
\\(35)! &amp;\equiv  -14\equiv 27 \mod 41 \\
\end{align}$ 

IS my work true ?
   can anyone says another idea ?
with a simple division, $$35!\equiv 27(mod41)$$
with a simple division, $$35!\equiv 27(mod41)$$