$$
237^{1002} = (23*10 + 7)^{1002} = \sum_{i=0}^{1001}23^{1002-i}10^{1002-i}\;7^i{1002 \choose i} + 7^{1002} =\\
[\textit{some huge honking multiple of }10] + 7^{1002} = \\
[\textit{some huge honking multiple of } 10] + 49^{501} = \\
[\textit{some huge honking multiple of }10] + (50 - 1)^{501} =\\
[\textit{some huge honking multiple of }10] + [\textit{some other gorfurshlugging multiple of }10] + (-1)^{501} =\\
[\textit{some huge honking multiple of }10] + [\textit{some other gorfurshlugging multiple of }10] - 1 = \\
[\textit{some huge honking multiple of }10] + [\textit{one less than some other gorfurshlugging multiple of }10] + 9
$$
So the last digit is $9$. Thing is only the last digits matter, and the last digits will cycle between $1, 7, 9, 3, 1, 7, 9, 3$. So you just need the last digit and the remainder of $1002$ divided by $4$.
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Crash course in modular arithmetic:
If you have some integer $N$ and we have two integers $a$ and $b$ so that $a = b \pm kN$ for some integer $k$ we say $a \equiv b \mod N$. We are basically considering an arithmetic system where we consider numbers by how much more than a multiple of $N$ they are.
Ex: If $4732895738927 \equiv 8647 \mod 10$ because $4732895738927 = 8647 + 10k$. Basically if $a \equiv b \mod 10$ then $a$ and $b$ have same last digit as $a = b + 10k$ for some $k$.
Lemma: if $a \equiv b \mod N$ and $c \equiv d \mod N$ then:
i) $a + c \equiv b+d \mod N$
ii) $ac \equiv bd \mod N$
iii) $a^n \equiv b^n \mod N$.
Pf: i) $a = b + kN$, $c = d + jN$ so $a+c = b + d + (j+k)N$ so $a+c \equiv b+d \mod N$.
ii) $a = b + kN$, $c = d + jN$ so $ac = (b+kN)(d+jN) = bd + (dk + bj)N + jkN^2 = bd + (dk + bj + jkN)N$. so $ac \equiv bd \mod N$.
iii) by induction $a^1 \equiv b^1 \mod N$ and if $a^n \equiv b^n \mod N$ then $a^{n+1} = a^na \equiv b^nb \mod N \equiv b^{n+1} \mod N$.
So we can apply this to your problem: $237 \equiv 7 \mod 10$ so $237^{1002} \equiv 7^{1002}$.
Notice: If you consider $0, 1,.....,N -1, N, 1 + N, ......, 2N-1, 2N, 2N + 1....$ there are at most $0,1,.....,N-1$ distinct values that can be equivalent $\mod N$ so for all the $a^k$ there must only a finite number of distinct things for $a^k$ to be equivalent $\mod N$ so there must be some $a^k \equiv a^j \mod N$ where $k \ne j$.
And if $a^k \equiv 1 \mod N$ then $a^{nk} = (a^k)^n \equiv 1^n \mod N \equiv 1 \mod N$.
So for example $7^2 \equiv 49 \equiv 9 \mod 10$
$7^3 = 7^2*7 \equiv \mod 10 \equiv 9*7 \equiv 63 \equiv 3 \mod 10$
$7^4 = 7^3*7 \equiv 3*7 \equiv 21 \equiv 1 \mod 10$.
So $7^{1000} = (7^4)^{250} \equiv 1^250 \equiv 1 \mod 10$.
Putting this all together:
$237^{1002} \equiv 7^{1002} = 7^{1000}*7^2 \equiv (7^4)^{250}*49 \equiv 1^{250}*49 \equiv 1*49 \equiv 9 \mod 10$
So $237^{1002}$ and $9$ have the same last digit; $9$.
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A theorem that is beyond this crash course is Euler's Thereom. If $N$ and $a$ have not common factors, and if $\phi(N) = $ the number of numbers $1,2, ....,N$ that have no common factors with $n$... then $a^{\phi(N)} \equiv 1 \mod N$.
So in your problem $\phi(10) = 4$ because $1,3,7, 9$ have no factors in common with $10$ while $2,4,5,6,8,10$ do. And $7$ and $10$ have no common factors... So $7^4 \equiv 1 \mod 10$. (And we can test that and $7^4 = 49*49 = 40^2 + 2*9*40 + 9^2 \equiv 81 \equiv 1 \mod 10$.)
So $237^{1002} \equiv 7^{1002} \equiv (7^4)^{250}7^2 \equiv 1^{250}49 \equiv 9 \mod 10$.
Best Answer
First of all, finding last $n$ digits in Base $10$ essentially finding modulo $\displaystyle10^n$
Using Carmichael function, $\displaystyle3^{\lambda(100)}\equiv1\pmod{100}$ as $(3,100)=1$
$$\implies3^{3^{2014}}\equiv3^{3^{2014}\pmod{\lambda(100)}}\pmod{100}$$
Now $\displaystyle\lambda(100)=20,$ so we need to find $\displaystyle3^{2014}\pmod{20}$
and by observation $\displaystyle3^4=81\equiv1\pmod{20}$ or using Carmichael function $\displaystyle\lambda(20)=4\implies3^4\equiv1\pmod{20}$
and as $\displaystyle2014\equiv2\pmod4\implies3^{2014}\equiv3^2\pmod{20}\equiv9$
$$\implies3^{3^{2014}}\equiv3^9\pmod{100}$$
Now $\displaystyle3^9=3\cdot3^8=3\cdot(3^2)^4=3(10-1)^4=3(1-10)^4=3\left[1-\binom4110+\binom4210^2-\binom4310^3+10^4\right]$
$\displaystyle\equiv3\left(1-40\right)\pmod{100}\equiv-117\equiv83$