In F41* = (Z/41Z)* determine 17^(-12).

I deduced that 17^(-12) is congruent to (17^(-1))^12 and congruent to 29^12 (mod 41), since 17*29 = 493 = 12*41 + 1 which is congruent ti 1 (mod 41)

After that I can't find a way to reduce 29^12 (modulo 41)

Any help would be appreciated!!!