Is It Simple..............

Hey just go with the power of $\huge{mod}$

(3+1) = 4 (mod 10)

( $5^{2}$ +1) = 6 (mod 10) (since, any power of 5 will have unit digit as 5 and 5+1=6)

Then we have

( $3^{3}$ +1) = ( $3^{2}$ )3 + 1 (mod 10)

( $3^{3}$ +1) = (-1)3 + 1 (mod 10)

( $3^{3}$ +1) = -2 or 8 (mod 10)

again we have,

( $5^{4}$ +1) = 6 (mod 10)

( $3^{5}$ +1) = ( $3^{2}$ )( $3^{2}$ )3 + 1 (mod 10)

( $3^{3}$ +1) = (-1)(-1)3 + 1 (mod 10)

( $3^{3}$ +1) = 4 (mod 10)

Multiplying all these we get,

$4 \times6 \times8 \times6 \times4$ (mod 10)

$4\times4$ =6 and $6\times6$ = 6 (mod 10)

6*8 = 8(mod 10)

Thus the unit digit is 8.

$\huge{:P}$

( ${ 5 }^{ 2 }$ +1)( ${ 5 }^{ 4 }$ +1)= 26 × 626

∴ Units digit of the product of ( ${ 5 }^{ 2 }$ +1)( ${ 5 }^{ 4 }$ +1) is 6

(3 + 1)( ${ 3 }^{ 3 }$ +1)( ${ 3 }^{ 5 }$ +1)=4 × 28 × 244

∴ Units digit of the product of (3 + 1)( ${ 3 }^{ 3 }$ +1)( ${ 3 }^{ 5 }$ +1) is 8

Now, the units digit of the product of (3 + 1)( ${ 3 }^{ 3 }$ +1)( ${ 3 }^{ 5 }$ +1)( ${ 5 }^{ 2 }$ +1)( ${ 5 }^{ 4 }$ +1) is same as the units digit obtained by multiplying 6 and 8.

Thus, the units digit of (3 + 1)( ${ 3 }^{ 3 }$ +1)( ${ 3 }^{ 5 }$ +1)( ${ 5 }^{ 2 }$ +1)( ${ 5 }^{ 4 }$ +1) is 8