<s>[INST] what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by 47? Can any one please tell the approach or solve the question 

what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by $47$?

I can solve remainder of $45!$ divided by $47$ using Wilson's theorem but I don't know what must be the approach for this model problems, as $47$ is a prime number I cannot convert it into another factorial and divide.
If any one of you viewing have any idea regarding the approach, please post your approach here.
Thanks in advance.
Regards,
Pavan Kumar
 [/INST] Just to compose table:
\begin{array}{|c|r|}
\hline
n! & \equiv \ldots (\bmod \:47) \\
\hline \\
1! & 1 \\
2! & 2\cdot 1 = 2 \\
3! & 3 \cdot 2 = 6 \\
4! & 4 \cdot 6 = 24 \\
5! & 5 \cdot 24 = 120 \equiv 26 \\
6! & 6 \cdot 26 = 156 \equiv 15 \\
7! & 7 \cdot 15 = 105 \equiv 11 \\ 
\cdots \\
44! & 44 \cdot 8 = 352 \equiv 23 \\
45! & 45 \cdot 23 = 1035 \equiv 1 \\
\hline
\end{array}
$45$ steps/rows in total.
Then to find sum:
$S = 1+2+6+24+26+15+11+\ldots+23+1 = \color{#E0E0E0}{1052 \equiv 18 (\bmod \: 47)}$.

Here we use idea: 
if $\qquad$ $k! \equiv s (\bmod \: p)$, 
then $\;$ $(k+1)! \equiv (k+1)\cdot s (\bmod \: p)$,
and apply it step-by-step.
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