To find the last digit of $2^{80}$, we need to find $2^{80} \text{ mod }10$.
Note that $\text{gcd}(2^{80}, 10)=2$, and $m \cdot a \text{ mod } m \cdot n = m (a \text{ mod } n)$. Then $2^{80} \text{ mod }10 = 2 \cdot (2^{79} \text{ mod }5)$.
Also, since $\text{gcd}(2,5)=1$ and $\phi(5)=4$, where $\phi$ is the Euler's Totient function, we have $2^4 \text{ mod } 5 =1$, by Euler's Totient Theorem.
Since $2^4 \text{ mod } 5 =1$, we have $2^{79} \text{ mod }5 = ((2^4)^{19} \cdot 2^3 ) \text{ mod }5 = 1\cdot3=3$.
Thus, $2^{80} \text{ mod }10 = 2 \cdot (2^{79} \text{ mod }5) = 2 \cdot 3 =6$, which is the last digit. Therefore, the answer is Option C.
Similarly, to calculate last two digits we need $2^{80} \text{ mod }100$, for last three digits we need $2^{80} \text{ mod }1000$, and so on.