$3$ , $4$ tower

$\begin{aligned} 3^1&=3\\3^2&=9\\3^3&=27\\3^4&=81\\3^5&=243\\3^6&=729\\3^7&=2187\\3^8&=6561\\3^9&=19683\end{aligned}\\\vdots\quad \vdots\quad \vdots$

Here we can see, changing exponent $4$ , the last digit remains same. Again if exponent is divisible by 4 than the last digit come $1$ .

Here, the $3$ 's exponent $\huge {\color{#20A900}{4}^{\color{#D61F06}{5}^{\color{#69047E}{6}^{\color{#624F41}{7}^{\color{#E81990}{8}}}}}}$ is divisible by $4$ .

So, it's last digit is $\boxed1$

Let the given number be $N = 3^{2n}$ , where $n$ is a positive integer. Note that the assumption is valid. Then we have:

$N \equiv 3^{2n} \equiv 9^n \equiv (10-1)^n \equiv (-1)^n \text{ (mod 10)}$

Note that $n$ is even and we have $N \equiv \boxed{1}\text{ (mod 10)}$ .

Let (n) end = last digit of a big number n.

(If there is an actual symbol for this, I don't know, so don't blame me)

Know:

3^1 end = 3

3^2 end = 9

3^3 end = 7

3^4 end = 1

4 raised to any number is always divisible by 4.

So, the exponent of 3 (overall) is divisible by 4.

Since it is always divisible by 4, 3^4...^8 will end in 1.

The answer is 1.

12/9/2017

Edit: Credit goes to Munem Sahariar for pointing this out. :)

4 raised to 0 is 1.

But we have a different case here.

4 raised to any positive integer greater than 1 is always divisible by 4.

Clearly, the exponent of 4, which is 5^6^7^8, is a positive integer greater than 1.

So, it is still the same.

3^4×5×6×7×8 = 3^6720 , 6720 is divisible by 4 so ans=1 because ( odd no.^4n = .............1 )

Starting from top 7^8 and 7 repeats its last digit after every 4 powers. As, 1,7,49,343,2401 And since we have 7^8 hence on counting 8, we get 7,9,3,1,7,9,3,1 So 1st part will last with a '1'

Then 6^(....1)=(....6)

5^(...6) = (....5) As 5 always results 5 as the last term of each of its power greater than 0

Then 4^5 = (....4) use the same rule

And then 3^4=(....1) as it repeats after every 4

Hence answer is 1