Rewrite the product as
[tex]13^4\cdot 17^2\cdot29^3 = (10+3)^4 (10+7)^2 (20+9)^3[/tex]
Powers of 10 and multiples of 10 don't contribute to the units digits, so we only need to consider the product
[tex]3^4\cdot7^2\cdot9^3=81\cdot49\cdot729[/tex]
and we can treat this the same way:
[tex]81\cdot49\cdot729=(80+1)(40+9)(720+9)[/tex]
Again, ignore the multiples of 10 to reduce this to the product 1*9*9 = 81, whose units digit is 1.
