1) I use RPF here for “Relatively Prime to Five.”

2) N > 1
3) For all divisions “/”, only quotient is considered. Fractions ignored.

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Solution:
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Let
P_1 = N/5, (Only quotient)
P_2 = N/(5^2)
P_3 = N/(5^3) and so on till .... ……. P_k = N/(5^k)
Beyond (5^k), higher powers of 5 gives quotient 0.

Let summation of Pi = P

Notice that {N!/(10)^P}(Mod 10) is what we need.
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Let RPF(q) indicate the product of first q natural numbers, which are relatively prime to 5.
Let RemRPF(q) indicate the reminder(when divided by 10) of first q natural numbers, which are relatively prime to 5.

Example RPF(11) = (1*2*3*4*6*7*8*9)*(11*12*13) ---------------------------------------------------------------------------(1)
It can be easily observed that RemRPF(q) is repetitive from RemRPF(9) onwards.

For example 
{ RemRPF(9) , RemRPF(10)….. RemRPF(16)}

= { RemRPF(17) , RemRPF(18)….. RemRPF(24)}

= {6,2,6,4,4,8,4,6} ----------------------------------------------------------------------------------------------------------------------(2)

Since it is repetitive, from (2), it can be seen that 

[RemRPF(q)] = [RemRPF(q(Mod 8))] -----------------------------------------------------------------------------(3)

Note that in (3), treat RemRPF(1) = 6, instead of 1. And also RemRPF(0) = RemRPF(8).


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2*2*2*2*2*2*2*2
The corresponding reminders at each stage of the above multiplication, are 

{2,4,8,6,2,4,8,6}

And it can be seen that this reminder pattern {2,4,8,6} repeats. ----------------------------------------------------------------(4)

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Let N(i) = RPF(4*P_{i+1} + ((N( Mod 5^{i+1}))/(5^i)} 

Note that 

(N!/10^P)(Mod 10) 

= [{\prod_{i=0}^k[N(i) ]}/2^P] (mod 10) 

= [[\prod_{i=0}^k{RPF(4*P_{i+1} + ((N( Mod 5^{i+1}))/(5^i)) }]/[2^P]](mod 10)---------------------------------(5)

Applying (2), (3) & (4) to (5) we get

(N!/10^P)(Mod 10) 

= [[\prod_{i=0}^k{RemRPF[RPF((4*P_{i+1} + ((N( Mod 5^{i+1}))/(5^i)))(mod 8) }]/[2^{P(mod 4)}]](mod 10)---------------------------------(6)


The expression in equation (6) can be easily calculated manually. In the numerator, each RemRPF[RPF((4*P_{i+1} + ((N( Mod 5^{i+1}))/(5^i)))(mod 8) } can be reduced to a single digit number.


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Example : N = 93
i = 0
= RemRPF[RPF((4*18 + ((93( Mod 5))/(5^{0})))(mod 8) }
= RemRPF[RPF((4*18 + 3)(mod 8)) }
= RemRPF[RPF(3)]
= 6
&&&&&&&&&&&&&&&
I = 1
= RemRPF[RPF((4*3 + ((93( Mod 25))/(5)))(mod 8) }
= RemRPF[RPF((4*3 + 3)(mod 8) }
= RemRPF[RPF(7)]
= 4
&&&&&&&&&&&&&&&&
I = 2
= RemRPF[RPF((4*0 + ((93( Mod 125))/(25)))(mod 8) }
= RemRPF[RPF((4*0 + 3)(mod 8) }
= RemRPF[RPF(3)]
= 6
&&&&&&&&&&&&&&&&&&
2^P (mod 10) = 2^21 (mod 10) = 2^1 (mod 10) = 2

The answer for 93! = ((6*4*6)/2)(mod 10) = 2