Last 3 digits of 7 $^{77}$

I don't know of any fancy, quick equation to assist with this, but I have a rather fun solution.

Whenever one is dealing with finding digits from powers, establishing a pattern is the first step to take. Listing out the powers of 7 mod 1000 (to limit them to 3 digits) we get:

$7, 49, 343, \textbf{401}, 807, 649, 543, \textbf{801}, 607, 249, 743, \textbf{201}, \dots$

If we look at every 4th term, a small pattern is emerging. Specifically it always ends with $01$ and leads with its current power placement mod 10 (Ex. $7^4=\dots401, 7^8=\dots801, 7^{12}=\dots201$

Now we see that $77\mod4 = 1$

So our solution, broken down, is $7^{77}=7^{76}\times7=\dots601\times7=\dots207$

Thus the answer is $207$

$\begin{aligned} 7^3N & \equiv 7^{80} \equiv 7^{2\times 40} \equiv 49^{40} \equiv (50-1)^{40} \equiv 1 \text{ (mod 1000)} \\ \implies 343N & \equiv 1 \text{ (mod 1000)} \end{aligned}$

Let $N \equiv \overline{abc}$ , a three-digit integer. Then,

$\begin{aligned} 343(100a + 10b + c) & \equiv 1 \text{ (mod 1000)} \\ 34300a + 3430b + 343c & \equiv 1 \text{ (mod 1000)} & \small \color{#3D99F6} \text{For last digit of LHS }=1 \implies c = 7 \\ 34300a + 3430b + 2401 & \equiv 1 \text{ (mod 1000)} & \small \color{#3D99F6} \text{For 2nd last digit of LHS }= 0 \implies b = 0 \\ 34300a + 2401 & \equiv 1 \text{ (mod 1000)} & \small \color{#3D99F6} \text{For 3rd last digit of LHS }= 0 \implies a = 2 \\ 68600+2401 & \equiv 1 \text{ (mod 1000)} \\ \implies N & \equiv \overline{abc} = \boxed{207} \end{aligned}$

$7^4 \equiv{401}\pmod{1000}$

$7^{77}\equiv{7^{76}\cdot 7}\equiv{{2401}^{19}\cdot 7}\equiv{{401}^{19}\cdot 7}\equiv{(400 + 1)^{19}\cdot 7}\equiv{(19\times{400} + 1)\cdot 7}\equiv{207}\pmod{1000}$

@Chew-Seong Cheong @Seth Christman Please comment!