We need the maximum n such that
(n+1)!-1 le 10^9 or
(n+1)! le 10^9+1. We will try to obtain a reasonable firts approximation for n.
Applying log to both sides
log((n+1)!) le log(10^9+1) or
sum_(k=0)^nlog(k+1) le log(10^9+1) or
int_0^n log(xi+1)d xi = (n+1)log(n+1)-n le log(10^9+1)
Now using an iterative procedure like Newton
x_(k+1)=x_k-f_k//((df)/(dx))_k with
f(x)= (x+1)log(x+1)-x - log(10^9+1)
we obtain in few iterations the value for n
n = floor11.75 = 11 so with n=11 we have
(11+1)!-1 = 479001599 lt 10^9 and
(12+1)!-1=6227020799 gt 10^9
