zsigmondy's ===> primitive prime factor of (p - 1)^p + 1 (p | (p - 1) + 1) alternatively assume ftsoc not, note (p - 1)^p + 1 is odd, so then (p - 1)^p + 1 is a power of p, a contradiction by catalan's another alternative approach is that if $(p - 1)^p + 1 = p^{\alpha}$, we have $v_p((p - 1)^p + 1) = 2$ by lte, so $\alpha = 2$, a contradiction as $(p - 1)^p > p^2$ (also these are prolly extremely overkill approaches)