The question asks for the "angka satuan" or "unit digit" of $777^{777}$.

To solve this, we need to find the last digit of $777^{777}$. However, we can simplify this by focusing on the last digit of the base 777, which is 7. This reduces the problem to finding the last digit of $7^{777}$.

Step-by-step approach:

1. The last digits of powers of 7 repeat in a cycle:

   7^1 &\equiv 7 \pmod{10}, \\ 7^2 &\equiv 49 \equiv 9 \pmod{10}, \\ 7^3 &\equiv 343 \equiv 3 \pmod{10}, \\ 7^4 &\equiv 2401 \equiv 1 \pmod{10}. \end{aligned}$$ So, the cycle of the last digits of powers of 7 is: $$7, 9, 3, 1$$.

2. The length of this cycle is 4. To find the last digit of $7^{777}$, we determine where $777$ falls in the cycle by finding $777 \mod 4$: $777 \div 4 = 194 \text{ remainder } 1.$ This means $777 \equiv 1 \pmod{4}$.

3. Since $777 \equiv 1 \pmod{4}$, the last digit of $7^{777}$ is the same as the last digit of $7^1$, which is 7.

Thus, the last digit of $777^{777}$ is 7.

The correct answer is C. 7.

Would you like more details or have any other questions?

Here are some related questions to expand your understanding:

1. What are the unit digits of $5^{10}$ and $6^{15}$?
2. How does modular arithmetic help in finding the last digits of large powers?
3. What is the unit digit of $3^{1234}$?
4. Can the unit digit of a number ever change if we add multiples of 10?
5. What patterns can we observe in the last digits of powers of 2?

Tip: Whenever solving powers for the last digit, observe if the digits form a repeating cycle; this simplifies the problem greatly!