Unit digit

Let $P = 2^{3^{17}}\times 3^{5^{25}}\times 4^{31}\times 7^{7^{93}}\times 8^{31^{32}}\times 19^{17} = A \times B \times C \times D \times E \times F$ .

Since 3, 7, 19 are 10 are coprime integers, for factors $B$ , $D$ and $F$ of $P$ , we can use Euler's theorem and Euler's totient function $\phi(10) = 10 \times \frac 12 \times \frac 45 = 4$ as follows:

$\begin{aligned} B & \equiv 3^{5^{25}} \equiv 3^{5^{25} \text{ mod 4}} \equiv 3^{(4+1)^{25} \text{ mod 4}} \equiv 3 \text{ (mod 10)} \\ D & \equiv 7^{7^{93}} \equiv 7^{7^{93} \text{ mod 4}} \equiv 7^{(8-1)^{93} \text{ mod 4}} \equiv 7^{-1 \text{ mod 4}} \equiv 7^3 \equiv 3 \text{ (mod 10)} \\ F & \equiv 19^{17} \equiv (20-1)^{17} \equiv -1 \equiv 9 \text{ (mod 10)} \end{aligned}$

As 2, 4 and 8 are not coprime integers with 10, we cannot use Euler's theorem, but the fact that $2^n \text{ mod 10} = \begin{cases} 6 & \text{if } n \text{ mod 4} = 0 \\ 2 & \text{if } n \text{ mod 4} = 1 \\ 4 & \text{if } n \text{ mod 4} = 2 \\ 8 & \text{if } n \text{ mod 4} = 3 \end{cases}$

$\begin{aligned} A & \equiv 2^{3^{17}} \equiv 2^{3^{17} \text{ mod 4}} \equiv 2^{(4-1)^{17} \text{ mod 4}} \equiv 2^{-1 \text{ mod 4}} \equiv 2^3 \equiv 8 \text{ (mod 10)} \\ C & \equiv 4^{31} \equiv 2^{2\times 31 \text{ mod 4}} \equiv 2^2 \equiv 4 \text{ (mod 10)} \\ E & \equiv 8^{31^{32}} \equiv 2^{3\times 31^{32} \text{ mod 4}} \equiv 2^{3\times (32-1)^{32} \text{ mod 4}} \equiv 2^3 \equiv 8 \text{ (mod 10)} \end{aligned}$

Therefore, we have:

$\begin{aligned} P & \equiv A \times B \times C \times D \times E \times F \text{ (mod 10)} \\ & \equiv 8 \times {\color{#3D99F6}3} \times 4 \times {\color{#3D99F6}3} \times 8 \times 9 \text{ (mod 10)} \\ & \equiv (-2) \times 4 \times (-2) \times {\color{#3D99F6}9} \times 9 \text{ (mod 10)} \\ & \equiv (-2) \times 4 \times (-2) \times {\color{#3D99F6}(-1)} \times (-1) \text{ (mod 10)} \\ & \equiv 16 \equiv \boxed{6} \text{ (mod 10)} \end{aligned}$