Find the remainder when $7^{7^{7}}$ is divided by 1000 I need help with this problem please 

Find the remainder when $7^{7^{7}}$ is divided by $1000$

My try follow
$1000=8×125$ , now 
$7 \equiv -1 \;\bmod\; (8)$ 
$\to$  $7^{7^{7}} \equiv -1 \;\bmod\; (8)$      and 
$7^{100}\equiv 1 \;\bmod\; (125)$
Any help to complete this solution? 
 A: Since $7^4=2\,401\equiv1\pmod{400}$ and since $7^3=343$, $7^7\equiv343\pmod{400}$. So, by Fermat-Euler, $7^{7^7}\equiv7^{343}\pmod{1\,000}$.
Since $7^4\equiv401\pmod{1\,000}$, $7^{20}=(7^4)^5\equiv1\pmod{1\,000}$. Therefore$$7^{343}=7^{340+3}\equiv7^3=343\pmod{1\,000}.$$
A: As $7^4=2400+1$  and $7\equiv-1\pmod4\implies 7^7\equiv-1\equiv3\pmod4$
We can write $7^7=4m+3$ where integer $m\ge0$
$$7^{4m+3}=7^3(1+2400)^m$$
Now $(1+2400)^m\equiv1+\binom m12400\pmod{1000}$
$7^8=(1+2400)^2\equiv1\pmod{40}\implies7^7\equiv7^{-1}\equiv23$
We can choose $7^7=40n+23=4(10n+5)+3$
$\implies m\equiv5\pmod{10}, m=10r+5$(say)
$1+2400m\equiv1+2400(10r+5)\equiv1\pmod{1000}$
A: $(7^{100} \mod 125)= 1$ so $7^{7^7} = 7^{(7^7 \bmod 100)} \mod 125 = 7^{43} \mod 125 = 93$
Now you can use Chinese remainder theorem with $x = 93 \pmod {125}$ and $x = -1 \pmod 8$, fastest way by hand is to check the $8$ values of $125 + 93k$ in the range of $0$ to $1000$ to see which one is $-1$ of $\pmod 8$.

Alternatively
$$\begin{align}
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7 \text{^} (7 \text{^}  7) \mod 1000 & = 
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7 \text{^} (7 \text{^}  7 \mod \Phi 1000) \mod 1000 \\
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&= 7 \text{^} (7 \text{^}  7 \mod 400) \mod 1000 \\
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&= 7 \text{^} 343 \mod 1000 \\
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&= 343 \\
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\end{align}$$
Just calculating $7 \text{^} 823543 \pmod {1000}$ is trivial with modular exponentiation as well.
