${\rm mod}\ 7\!:\ \color{#c00}{4^{\large 3}\equiv 1}\,\Rightarrow\, 4^{\large 1000}\equiv 4^{\large 1+999}\equiv 4 (\color{#c00}{4^{\large 3}})^{\large 333}\equiv 4\color{#c00}{(1)}^{\large 333}\equiv 4$
More generally we have that $\ 4^{\large r+3q}\equiv 4^r (\color{#c00}{4^{\large 3}})^{\large q}\equiv 4^{\large r}\color{#c00}{(1)}^{\large q}\equiv 4^{\large r}$
Written in terms of mod this is: $\ 4^{\large n}\equiv 4^{\large n\ {\rm mod}\ 3}\,$ where $\ n = 3q+r\,$ and $\,r = n\ {\rm mod}\ 3$
I don't want to bury you in notation, so I will settle for examples and I hope you will see the implied pattern.
\begin{align}
1234567\pmod 9
&= (1+2+3+4+5+6+7) \pmod 9\\
&= 28 \pmod 9\\
&= (2+8) \pmod 9\\
&= 10 \pmod 9\\
&= (1+0) \pmod 9\\
&= 1\\
\end{align}
\begin{align}
1234567\pmod{99}
&= (1+23+45+67) \pmod{99}\\
&= 136 \pmod{99}\\
&= (1 + 36) \pmod{99}\\
&= 37\\
\end{align}
\begin{align}
1234567\pmod{999}
&= (1+234+567) \pmod{999}\\
&= 802
\end{align}
The verification of that last computation would look like this
\begin{align}
1234567
&= 1 \times 1000000 + 234 \times 1000 + 567 \pmod{999}\\
&= 1 \times (1 + 999999) + 234 \times (1 + 999) + 567 \pmod{999}\\
&= (1 + 999999) + (234+ 234 \times 999) + 567 \pmod{999}\\
&= 1 + 234 + 567 \pmod{999}\\
\end{align}
So
\begin{align}
123,123, \ldots ,123 \pmod{999}
&= 123+123+\ldots + 123 \pmod{999}\\
&= 100 \times 123 \pmod{999}\\
&= 12300 \pmod{999}\\
&= (12 + 300) \pmod{999}\\
&= 312
\end{align}
For the case $10^n + 1$, it looks like this
\begin{align}
1234567\pmod{11}
&= (7-6+5-4+3-2+1) \pmod{11}\\
&= 4 \pmod{11}\\
&= 4\\
\end{align}
\begin{align}
1234567\pmod{101}
&= (67 - 45 + 23 - 1) \pmod{101}\\
&= 44\\
\end{align}
\begin{align}
1234567\pmod{1001}
&= (567-234+1) \pmod{1001}\\
&= 334
\end{align}
The verification of that last computation would look like this
\begin{align}
1234567
&= 1 \times (1001000 - 1001 + 1) + 234 \times (1001-1) + 567 \pmod{1001}\\
&= 1 - 234 + 567 \pmod{1001}\\
&= 1 \times (-1001 + 1) + 234 \times (-1) + 567 \pmod{1001}\\
&= 567 - 234 + 1 \pmod{1001}\\
&= 334\\
\end{align}
Note similarly that
$$10^9 = 1000000000 = 1001000000 - 1001000 + 1001 - 1 = -1 \pmod{1001}$$
Best Answer
For $k \geq 7$, we have $$7\; \vert\; k!.$$ Hence $$\sum_{k=1}^{49} k! = 1!+2!+3!+4!+5!+6!\equiv1+2+(-1)+3+1+(-1)\equiv5 \pmod{7}.$$