$14$ all the way ...

The last digit of ${{{{14^{14}}^{14}}^{14}}^{\cdots}}$ is $\boxed{6}$ because $14^{1} = 4, 14^{2} = 6, 14^{3} = 4, 14^{4} = 6, \cdots$ . Thus, the answer must be $\boxed{6}$ .

$14\upuparrows 14\quad means\quad { 14 }^{ { 14 }^{ { 14 }^{ 14{ . }^{ . } } } }.$ Clearly, if we look at the occurrence pattern of $14$ , we see that if the power of $14$ is even, the unit's digit is $6$ and if the power is odd, the unit's digit is $4$ .

From the question, the power of the first base $14$ , is clearly even. So, the unit's digit will definitely be $6$ .

$14 \uparrow\uparrow 14 = 14^{14 \uparrow\uparrow 13}$

The exponent is thus $14^a$ where $a$ is an integer greater than 1. Hence, $14^a = 0 \mod 4$

Therefore, $14^{14^a} \mod 10 = 4^4 \mod 10 = \boxed{6}$