To calculate remainder for a base having 2 exponents, You can follow the below shortcut method...
Remember 32^32^32 is not equal to 32^(32*32) ...

Here 32^32^32 % 9
(9*4-4^32^32)%9
(-4)^32^32

Base 4 nd divisor 9 are co-primes... So we can use Euler's Theorem...

Euler's Theorem :
*****************
R[M^N % P] ... If M nd N are co- primes , them We can use this theorem...
We have to calculate E(N) = N*(1-1/a)*(1-1/b)...
For example if i want to calculate E(15),
15 = 3^1*5^1
Then E(15) = 15*(1-1/3)*(1-1/5) => E(15) = 8
Similarly 9=> 3^2
E(9) = 9*(1-1/3) = 6
Then R[M^E(N) % P] = 1 [This is the Euler's Form]

So now we have to calculate (-4)^32^32 % 9
If we have 2 exponents (R[M^N^O] % P], this is the procedure
Calculate N^O % E(N) = x
Then Calculate M^x % P

So 32^32 % 10 => [10*3 + 2]^ 32 % 6
2^32 % 6
We can use cyclic rem. theorem here...
2^1 % 6 = 2
2^2 % 6 = 4
2^3 % 6 = 2 [Again]
2^4 % 6 = 4
So cyclicity = 2 [upto 2nd power]So rem ll be 2k+n
So 32 => 2*15+2 [k=15,n=2]
If n=1, rem=2
If n=2, rem=4
So rem[2^32 % 6] = 4

Now Calculate (-4)^4 % 9
=> 256 % 9
= 4%9
Hence rem is 4

Am sorry...
32^32%10=(10*3+2)^32%6 .... Its a mistake...

If we have 2 exponents (R[M^N^O] % P], this is the procedure
Calculate N^O % E(N) = x
Then Calculate M^x % P

So 32^32 % 6 => [6*5 + 2]^ 32 % 6 [Here E(N) = 6... N^O = 32^32]
Rem[32^32 % 6] = 4

And Yes 32=> 2^5
So E(32) = 32*(1-1/2) = 16

1 will be the remainder

First of all we have to find out the unit digit of 32^32 , which is 6
It means (32^6)/9 will give us the remainder.

Now, (32^6)/9 = (9*3+5)^6 /9
= (5)^6 / 9
= {(5)^3}^2 / 9
= (125)^2 / 9
= (9*14-1)^2 / 9
= (-1)^2 / 9
= 1 / 9
= 1

32^1 % 9 = 5
So (5^32^32) % 9
Euler value of(9) = 6
((5^6)^5 * 5^2)^32
5^64 % 9
E(9) = 6
(5^6)^10 * 5^4 % 9
625 % 9 = 4

Ans : 4

Saraswathy answer is 100 percent correct.
Can you please explain the method to solve....?
How Euler Value of (9) is 6 and how u put the values in form:
((5^6)^5 * 5^2)^32

Plzzz....and thx for ur reply !!

Euler's Theorem :
*****************
R[M^N % P] ... If M nd P are co- primes , them We can use this theorem...

@Saraswathy
Pls explain the statementa given below
32^32%10=(10*3+2)^32%6

I mean E(32)=32*(1-1/2)=16
Right?
I am confused.

Thanks Saraswathy for ur detailed explanation....thank u very much !! :)

32^32^32/9
32/9=REM 5
5^32^32/9
5/9=5;25/9=7;125/9=8;5^4/9=4;5^5/9=2;5^6/9=1;5^7/9=5 REPEATING
32^32/6->32/6=2(REM)
2^32/6->2/6=2;4/6=4;8/6=2 REPEATING
32/2=0(REM);
2^32/6=1;
5^1/9=5
ANSWER 5