Now we are evolving

Let the number be $\large 97^{93^a}$ and we need to find $97^{93^a} \equiv x \text{ (mod 10)}$ . Since 97 and 10 are coprime integers, we can apply Euler's theorem .

$\begin{aligned} \large 97^{93^a} & \equiv \large 97^{93^a \text{ mod }\color{#3D99F6}{4}} \text{ (mod 10)} \quad \quad \small \color{#3D99F6}{\phi (10) = 4} \\ & \equiv \large 97^{(4 \times 23 + 1)^a \text{ mod 4}} \text{ (mod 10)} \\ & \equiv \large 97 \text{ (mod 10)} \\ & \equiv \large \boxed{7} \text{ (mod 10)} \end{aligned}$

Did same but I have a second method

83^{81} \equiv -1\ pmod{4}

So 87^{83}^{81} \equiv -1\pmod{4}

Proceeding same we got 93^{91}^{89}^{87}^{83}^{81} \equiv 1\pmod{4}

Now every power of 97 of form 4q+1 has unit digit 7