What is the units digit of 13^4*17^2*29^3?

I think that the fastest and more important, the safest way to solve this is:
First brake the problem to three multiplication problems:
1. 13*13*13*13
2. 17*17
3. 29*29*29
In all of the problems we don't have to multiply by the whole number, but only by the last digit (because that is what we are asked about).
Second, simply multiply: 13*13*13*13. It is actually 3*3*3*3 (we are interested in the last digit). Thus - 3*3=9, 9*3=7(the last digit), 7*3=1 (the last digit).
17*17 is actually 7*7 = 9 (the last digit).
29*29*29 is actually -9*9*9- that can be interpreted into 9*9=1 (the last digit) and 1*9=9.
So eventually we will have- 9 (29^3) * 9 (17^2) * 1 (13^4). Therefore - 9*9=1(the last digit), 1*1=1

So the correct answer is E -1.