$\gcd (2016!+1, 2015!+1)$ Can someone tell how to do this? I know the answer when there is no additional 1 in it. But with +1, I have no clue. Can someone give insights? I tried using $\gcd(a,b) = \gcd(a, b-a)$ but could not get anywhere. Thanks in advance

Since 
$$2016!+1=2016(2015!+1)-2015,$$
Euclid's algorithm yields
$$\gcd(2016!+1,2015!+1)=\gcd(2015!+1,2015)=1$$
