Let’s find the patterns of 11^n, 7^n, or 13^n.

The last digit of 11^n will always be 1, because we will always have 1*1 for the ones column multiplication.

For 7^n we should find a pattern for the ones column multiplication
7*7 = 49, 7*9 = 63, 7*3 = 21, 7*1 = 7, and we are back to 7*7. Like this:

7^0 = …1
7^1 = …7
7^2 = …9
7^3 = …3
7^4 = …1

So for 7^n:
If n mod 4 = 0, then the last digit is 1
If n mod 4 = 1, then the last digit is 7
If n mod 4 = 2, then the last digit is 9
If n mod 4 = 3, then the last digit is 3

Since 2004 mod 4 = 0, that means the last digit of 7^2004 is 1.

Similarly for 13^n… 3*3 = 9, 3*9 = 27, 3*7 = 21, 3*1 = 3, and we are back to 3*3. Like this:

13^0 = …1
13^1 = …3
13^2 = …9
13^3 = …7
13^4 = …1

So for 13^n:
If n mod 4 = 0, then the last digit is 1
If n mod 4 = 1, then the last digit is 3
If n mod 4 = 2, then the last digit is 9
If n mod 4 = 3, then the last digit is 7

Since 2005 mod 4 = 1, that means the last digit of 13^2005 is 3.

So the answer is 1*1*3 = 3.