Question

http://prntscr.com/4ny7yq

Answer 1

\[\large \binom{p-1}{k} = \dfrac{(p-1)!}{k!(p-k-1)!}\]

Answer 2

\[= \dfrac{(p-1)(p-2)\cdots (p-k)}{k!} \equiv \dfrac{(0-1)(0-2)\cdots (0-k)}{k!}\equiv -1 \pmod p\]

Answer 3

any other method without using that dumb formula ?

Answer 4

**** \[= \dfrac{(p-1)(p-2)\cdots (p-k)}{k!} \equiv \dfrac{(0-1)(0-2)\cdots (0-k)}{k!}\equiv (-1)^k \pmod p\]

Answer 5

oh well this one is good proof

Answer 6

that was just algebra
doesnt teach me anythng new

Answer 7

Ok moving to next question if u don't have anything to add :)

Answer 8

\[\large (a+1)^{p-1} = \sum\limits_{k=0}^{p-1} a^{p-1-k}\]

Answer 9

sounds more like binomial