A number theory problem by A Former Brilliant Member

$4^1=4$

$4^2=6$

$4^3=4$

$4^4=6$

The unit's digit repeat in every cycle of 2. We divide the exponent by 2. $\frac{39315}{2}=19657$ $remainder$ $1$ . Since $1$ is the remainder, the unit's digit is $4.$

$4^1 = \color{#D61F06}{4}$

$4^2 = 1\color{#D61F06}{6}$

$4^3 = 6\color{#D61F06}{4}$

$4^4 = 21\color{#D61F06}{6}$

Odd powers of $4$ end in $4.$

Even powers of $4$ end in $6.$

This applies to any base ending in $4.$

Since the power $39315$ is odd, $387934^{39315}$ ends in $4.$