Finding unit's digit in exponentiation Could someone please explain to me how to find the unit's digit in the following expression:
$$7^{95} - 3^{58}$$

Unit Digit means $\pmod{10}$
Observe that $7\equiv7\pmod{10},7^2=49\equiv9,7^3\equiv9\cdot7\equiv3,7^4\equiv3\cdot7\equiv1$
$3\equiv3\pmod{10},3^2=9\equiv9,3^3=27\equiv7,3^4=81\equiv1$
So, both $3,7$ have a cycle with period $=4$ which can be also confirmed using Euler's Totient Theorem with $\phi(10)=\phi(2)\phi(5)=4$ and $(3,10)=(7,10)=1$
As $95=23\cdot4+3$ and $58=4\cdot14+2,$
$7^{95}=(7^4)^{23}\cdot7^3\equiv1^{23}\cdot3\pmod{10}$
and $3^{58}=(3^4)^{14}\cdot3^2\equiv1^{14}\cdot9\pmod{10}$
