Power

$\begin{aligned} 2017^{2015} & \equiv 2017^{\color{#D61F06}{4} \times 503 + 3} \pmod{10} \\ & \equiv \color{#D61F06}{1}\times 2017^3 \pmod{10} \quad \quad \color{#D61F06}{[\text{See Note}]} \\ & \equiv (2020 - 3)^3 \pmod{10} \\ & \equiv (- 3)^3 \pmod{10} \\ & \equiv -7 \pmod{10} \\ & \equiv \boxed{3} \pmod{10} \end{aligned}$

$\color{#D61F06} {\text{Note: }}$ Since $2017$ and $10$ are coprimes therefore, we can apply Euler's totient theorem. $\Rightarrow 2017^{\phi (10)} \equiv 2017^4 \equiv 1 \pmod{10}$

take the last digit of 2017 that is 7. so now 7^2015 units digit will be the answer.

Units digit of powers of every number is periodical with period 4.

So, for 7 it is 7,9,3,1.

Now, 2015 = 4x503+3. so the third periodical units digit is the answer that is 3.

This process is not so good as the second last digit cannot be found.

Please suggest a better method.