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#
