First, $2003^n \equiv 3^n \mod 1000$.
$3$ is invertible modulo $1000$. The group of invertibles of $\mathbb{Z}/1000\mathbb{Z}$, $(\mathbb{Z}/1000\mathbb{Z})^\times$ has cardinality $\varphi(1000) = 1000 * 1/2 * 4/5 = 400$.
This implies that $3^{400} \equiv 1 \mod 1000$, and so $3^n \equiv 3^{n \mod 400} \mod 1000$.
So in order to comptute $2003^{2002^{2001}} \mod 1000$, we need to know $2002^{2001} \mod 400$.
$2002^n \equiv 2^n \mod 400$. This time, $2$ is not invertible modulo $400 = 2^4 * 25$.
For $n \geq 4$, $2^n$ is always a multiple of $2^4$, so $2^n \mod 400 = (2^4*2^{n-4}) \mod (2^4*25) = 2^4*(2^{n-4} \mod 25)$.
Now, $2$ is invertible modulo 25, and the group $(\mathbb{Z}/25\mathbb{Z})^\times$ has cardinality $\varphi(25) = 25*4/5 = 20$. This implies that $2^{20} \equiv 1 \mod 25$, and so $2^n \equiv 2^{n \mod 20} \mod 25$.
Putting all of this together, we get :
$2002^{2001} \mod 400 = 2^{2001} \mod 400 = 2^4 * (2^{1997} \mod 25) = 2^4 * (2^{1997 \mod 20} \mod 25) $
$=2^4 * (2^{17} \mod 25) = 2^4 * (131072 \mod 25) = 2^4 * 22 = 352$.
And finally $2003^{2002^{2001}} \mod 1000 = 3^{352} \mod 1000 = 241$.
The usual fast way is by modular arithmetic. For 10, the significant numbers are 2 and 5, which cube to give 8 and 125.
For $2$, all even numbers give a cube that is $0$ mod $8$. The odd numbers cube to themselves, modulo 8, so the valid answers modulo 8, are $0, 1, 3, 5, 7$, ie $5/8$ of the total.
For $5$, all multiples of $5$ cube to a multiple of $125$, but the remaining numbers are freely occurable. For example, there is a number whose cube is $3$, mod $125$. So for $5$, there are $100+1$ possible endings, or $101/125$.
Over $1000$, then there are $5/8 * 101/125$ = $505/1000$. For $100$, one reduces the case for 2 to modulo $4$, ie $0, 1, 3$, and for $5$, it's modulo $25$ (ie 21 solutions). There are thus $63=3*21$ possible endings for this.
When the fifth power is involved, a different thing happens. The powers of 5 must (in base 5), end in the digits $00,\; 01,\; 12,\; 33,\; 44$ only. The powers of $2$ are identical to above. So there are just $5$ possible endings for $25$ and $3$ possible endings for $4$, or just $15$ possible endings for $100$.
The ultimate form of this leads to the four cases where $10^a \mid x^n-x$, ie $-0001$, $-0000$, $-0625$ and $9376$, as the numbers $2$ and $5$ divide the value $x$.
When one handles a base with $3$ prime divisors, one does this for each of the three divisors.
Details
The process is weakly multiplicative. That means, if $x$ and $y$ are co-prime (ie $\operatorname{gcd}(x,y)=1$, then $f(xy)=f(x)\cdot f(y)$
Consider for example, the squares, modulo 20. This is weakly multiplicative, since we can write $f(20)=f(5) f(4)$, but not $f(20)=f(2) f(10)$. These are fairly easy to enumerate, since they all occur singly in the first six squares ($0$ to $5$).
0 1 4
0 1 2 3 4 5 0 0 16 4
0 1 4 9 16 5 = 6 1 5 1 9
Rows = mod 4, columns = mod 5
There are also 6 for modulo 15, which gives
0 1 4
0 1 2 3 4 5 6 0 0 6 9
0 1 4 9 1 10 6 1 10 1 4
Rows = mod 3, columns = mod 5
One sees that the modulo against $5$ does not depend on the other factors of the base. Moreover, there is a full column of the other primes for each case of 5. We can deal with the individual prime powers separately. So when we disect 10 or 14 below, instead of dealing with the full number, we can deal with the prime powers separately, and multiply the lot together.
One really has to worry if $\operatorname{gcd}(n, p^2-p) > 1$. In this case, there is a reduction of available choices.
For multiples of $p$, there is only one solution up to $p^n$ (ie -$0000000$ base $p$
When the gcd is $p$, then some $p^m$ divides $n$, and then there is only one string of $m+1$ digits, base $p$, for each primitive digit of $p$.
When the gcd ($x$) divides $p-1$, then there are just $(p-1)/x$ possible endings for non-multiples of $p$.
To give an example, suppose the power is $14$, and we're dealing with a base of 14. What is the count the available digits that $x^14$ might end in base $14$?
For prime $7$, the first rule tells us that all multiples of $7$ end in $00$ base $7$. This means that 1/7 of all numbers give rise to just 1 ending.
The second rule tells us that 7 divides $\operatorname{gcd}(14, 7^2-7)$. This means that there are just $7$ possible endings for 7th powers in base 7 (ie $00$, $01$, $24$, $25$, $42$, $43$ and $66$). We see that $6/7$ of the possible endings are removed, whether or not we count $00$: it's the same for each place.
The third rule tells us that 2 divides $\operatorname{gcd}(14, 7^2-7)$, so just half of the numbers (ie 3) are squares. These are $1$, $2$ and $4$.
All together, the modulo-7 restriction tells us that the only solutions for this problem, is that it is one of $1 + 42/7/2 = 4$ answers: here $00$, $01$, $24$ and $42$, base $7$
For $2$, the same rules tells us that there are just two answers over modulo $4$, that is, $00$ and $01$, base $2$
The chineses remainder solution can be used to find these $4*2=8$ solutions.
Best Answer
As we're interested in the last two digits, it suffices to compute this modulo $100$. Hence $$ 2017 ^{2017} \equiv 17^{17} \equiv 77 \pmod{100}. $$ The first equivalence holds by Euler's theorem (see the mothertopic mentioned by J. Lahtonen).