The previous solutions are not optimal. The complexity is exactly $n\log n + O(1)$ in calls to RandNum50 and is described in some detail here, using as a source of random bit (as suggested by Vor):
if ( rand50() > 25 ) then b = 1 else b = 0 // random bit
The basic idea is that you save a lot of bits if you generate a uniform between $1$ and $n!$, and then using factorial base decomposition, instead of generating a sequence of uniforms ranged up to $1$, then $2$, then $3$, etc., $n$. This is actually, as I mention in the post, the topic of a paper I have submitted!
If you do not know how to generate a uniform, as suggested in that post, from a random bit, you could also generate an approximation of the uniform directly, in this way (which is equivalent to Vor's "trulyrand", but faster):
P = (RandNum50()-1) + (RandNum50()-1)*50^1 + (RandNum50()-1)*50^2 + ...
going as far as you need to go. This is developing $P$ in base $50$. Then simply truncate $P$, i.e., $Q=P\mod n$, in your case $n=100!$. This value is not completely random, but it is a measure of uniformity that is often used. Or, as Vor suggests, you can reject if $P>n$. Then with this value, you can do the factorial base expansion as described in the post.