Calculators won't work!

$\begin{array}{l|c|c|c|c|c|c|c|c}\textrm{Power}&1&2&3&4&5&6&7&8& \\ \hline \textrm{Last digit}&3&9&7&1&3&9&7&1&\end{array}$

On observing the above data.

$\text{For, }3^n \text{, and n's divisors } \in 1, 2, 3, 4\\ \text{If highest divisor of n is 1, }3^n \text{, ends with 3} \\ \text{If highest divisor of n is 2, } 3^n \text{, ends with 9} \\ \text{If highest divisor of n is 3, } 3^n \text{, ends with 7} \\ \text{If highest divisor of n is 4, } 3^n \text{, ends with 1} \\ \therefore 3^{2004} \text{ends with 1, as 4 is the highest divisor of n, where divisor of n } \in 1, 2, 3, 4$

$\large{3^{2004}\equiv (3^4)^{501}\equiv 1^{501}\equiv\color{#20A900}{\boxed{\boxed{1}}}\pmod{10}}$

We can use the Euler Totient function to find the expression $\pmod{10}$ because $gcd(3,10)=1$ :

$\phi{(10)}=10 \times \left (1-\dfrac{1}{2} \right) \times \left (1-\dfrac{1}{5} \right) =4$

$3^{2004} \equiv 3^{2004 \pmod{4}} \equiv 3^0 \equiv \boxed{\boxed{1}} \pmod{10}$

There will be an infinite pattern when you always (x3) : 3,9,7,1,3,9,... (last digit). Knowing that $3^4$ and $3^8$ will have the last digit 1 , 3^2004 will also have the last digit 1. (Multiples of 4)

$2004 \equiv 0 ( \mod 4 ) \rightarrow 3 ^ { 2004 } = 3 ^ { 0 } = \boxed { 1 }$ .

${ 3 }^{ 1 }\equiv 3(mod10)\\ { 3 }^{ 2 }\equiv 9(mod10)\\ { 3 }^{ 3 }\equiv 7(mod10)\\ { 3 }^{ 4 }\equiv 1(mod10)$

${ 3 }^{ 2004 }\equiv { \left( { 3 }^{ 4 } \right) }^{ 501 }(mod10)\\ { 3 }^{ 2004 }\equiv { 1 }^{ 501 }(mod10)\\ { 3 }^{ 2004 }\equiv 1(mod10)$

So, the last digit is $1$ .