Last three digits of a large exponent, without using modular arithmetic

Question

If i have a really large number like \begin{gather} 3^{762259784987} \end{gather} or \begin{gather} 7^{836793257}\end{gather} Is there a way to find the last three digits without using modular arithmetic or is that the only method. If their is could someone please outline a method because i would be very curious. I don't want to use modular arithmetic, despite it probably being the most obvious method, because we haven't covered the topic yet but I still received a similar problem which leads me to believe that their could be another method that relies more on problem solving skills. But I could be very wrong and maybe the only method involves modular arithmetic. I worked out how to find the last digit because of the cyclical nature but I haven't had any luck with figuring out all of the last three digits.

(Note: Someone rightfully pointed out that the title and the question asked for two different things, my fault and sorry for any confusion but I was originally thinking about the last three)

Answer 1

Update: Last three digits of $3$ requires lots of work but not impossible as has already been in the comments. With the last three digits of $7$, we are a little bit luckier, though:

1 7 <-
2 49
3 343
4 401
5 807
6 649
7 543
8 801
9 607
10 249
11 743
12 201
13 407
14 849
15 943
16 601
17 207
18 449
19 143
20 1
21 7 <-

So, the last three digits are repeated per $20$ steps.

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Looking at the two last digits of powers of $3$ until the repetition, we get the following list:

1 3 <-
2 9
3 27
4 81
5 43
6 29
7 87
8 61
9 83
10 49
11 47
12 41
13 23
14 69
15 7
16 21
17 63
18 89
19 67
20 1
21 3 <-

So, the last two digits are repeated per $20$ steps.

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Looking at the two last digits of powers of $7$ until the repetition, we get the following list:

1 7 <-
2 49
3 43
4 1
5 7 <-

So, the last two digits are repeated per $4$ steps.

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