Verified answer

{(n+p)(p)} kongruent 1 mod p means

when dividing {(n+p)(p)} with p, there is a rest of 1

there exists an natural k, so that{(n+p)(p)}=k*p+1

n=1: {(1+p)(p)}=(1+p)!/1!p!=p!*(p+1)/p!=p+1, so k=1

n->n+1

let's say that there exists a k, so that {(n+p)(p)}=kp+1

now prove it for {(n+1+p)(p)}=mp+1

{(n+p+1)(p)}=(n+p+1)!/[(n+1)!p!]=

(n+p)!*(n+p+1)/[n!*(n+1)*p!]=

[(n+p)!/n!p!]*[(n+p+1)/(n+1)]=

{(n+p)(p)}*[(n+p+1)/(n+1)]=

(kp+1)*(n+p+1)/(n+1)=

(kpn+kp²+kp+n+p+1)/(n+1)=

[(kpn+kp²+kp+p) + (n+1)]/(n+1)=

p(kn+kp+k+1)/(n+1) + (n+1)/(n+1)=

p*[(kn+kp+k+1)/(n+1)]+1

because the first part has p as a factor, it can be divided with p without a rest, so it leaves only the 1 as rest

{(n+p+1)(p)}=mp+1, with m=(kn+kp+k+1)/(n+1)