Now my calculator needs a painkiller

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241

If the units digit of a number N is 3, then the units digit of N^2 will be 9, for N^3 it will be 7, for N^4 it will be 1 and then it will repeat (N^5 will be 3 again, N^6 will be 9 again and so on).

For the last two digits, it will repeat after 20 iterations (so if N^1 ends with 03, then N^21 also ends with 03 and if N^1 ends with 13, then N^21 also ends with 13). For the last three digits, it will repeat after 100 iterations.

So, now we need to find 2002^2001 mod 100 (since it repeats after the 100th iteration). Fortunately, we can employ a similar trick. The last two digits where are new number M ends in 2 will repeat after 20 iterations (with one small caution coming up), just like with the 3s. So, 2002^2001 is going to go through the cycle of 20 one hundred times and then it will be back to 1 again. Now, we can calculate that it should follow the pattern of the first number and end in 02. However, if you run this trial with 2s, we find that the first one ends with 02, but M^21 ends with 52 (M^2 and M^22 will both end in 04 and all the rest follow the same pattern except for this very first number which locks into 52 after the first iteration). So, the last two digits of 2002^2001 are 52.

Now, that means that the last three digits of 2003^2002^2001 will be the same as 3^52. Now, 3^12 will end with 441, 3^32 will end with 841 and then 3^52 will end with 241.