Your calculator would fail!

We note that any power of an integer ends with $6$ always ends with $6$ . Therefore, we have:

$\begin{aligned} 32^{593} & \equiv (30+2)^{593} \text{ (mod 10)} \\ & \equiv 2^{593} \text{ (mod 10)} \\ & \equiv 2 \cdot 16^{148} \text{ (mod 10)} \\ & \equiv 2\cdot 6 \text{ (mod 10)} \\ & \equiv \boxed 2 \text{ (mod 10)} \end{aligned}$

We just need the last digit, so we only use the last digit,i.e., $2$ .

The powers of $2$ continue in the pattern $2,4,8,6,\cdots$

Since $593\div 4$ gives remainder $1$

So, the last digit is $2^1=\boxed{2}$

Based on $2^n$ 's second digit = $2, 4, 8, 6, 2...$ :

$593 (\text{mod} (4)) = 1$

Substitute $n = 1$ - $2^1 = 2$

The answer is $\fbox 2$

32^593=(2^5)^593=2^2965. Then we cannot calculate it with a normal calculating method. Then, how we calculate it? Thankfully, there is a way to know the last digits of a few powers. But we need only 1 because we calculated it at the first, 32^593=2^2965. 2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64..... No way for these calculations for getting the correct answer if the exponent is a large number. In the latest calculation, we learned a few last digits of 2. The last digits are circuiting by the 4 places, like this, 2, 4, 8, 6. Then we can get the last digits of 2^2965. First, we divide the exponent by 4. Then we get 1. Find the last digits of 2^1. Then the answer to 5^593 is 2 because the last digit of 2^2965 is 2.