==> arithmetic/digits/least.significant/factorial.s <==
Reduce mod 10 the numbers 2..n and then cancel out the
required factors of 10. The final step then involves
computing 2^i*3^j*7^k mod 10 for suitable i,j and k. 

A small program that performs this calculation is appended. Like the
other solutions, it takes O(log n) arithmetic operations.

-kym
===

#include<stdio.h>
#include<assert.h>

int	p[6][4]={
	/*2*/	2,	4,	8,	6,
	/*3*/	3,	9,	7,	1,
	/*4*/	4,	6,	4,	6,
	/*5*/	5,	5,	5,	5,
	/*6*/	6,	6,	6,	6,
	/*7*/	7,	9,	3,	1,
	};

main(){
	int	i;
	int n;

	for(n=2;n<1000;n++){
		i=lsdfact(n);
		printf("%d\n",i);
		}

	exit(0);
	}

lsdfact(n){
	int	a[10];
	int	i;
	int	n5;
	int	tmp;

	for(i=0;i<=9;i++)a[i]=alpha(i,n);

	n5=0;
/* NOTE: order is important in following */
l5:;
	while(tmp=a[5]){	/* cancel factors of 5 */
		n5+=tmp;
		a[1]+=(tmp+4)/5;
		a[3]+=(tmp+3)/5;
		a[5]=(tmp+2)/5;
		a[7]+=(tmp+1)/5;
		a[9]+=(tmp+0)/5;
		}
l10:;
	if(tmp=a[0]){
		a[0]=0;	/* cancel all factors of 10 */
		for(i=0;i<=9;i++)a[i]+=alpha(i,tmp);
		}
	if(a[5]) goto l5;
	if(a[0]) goto l10;

/* n5 == number of 5's cancelled; 
   must now cancel same number of factors of 2 */
	i=ipow(2,a[2]+2*a[4]+a[6]+3*a[8]-n5)*
		ipow(3,a[3]+a[6]+2*a[9])*
		ipow(7,a[7]);
	assert(i%10);	/* must not be zero */
	return	i%10;
	}

alpha(d,n){
/* number of decimal numbers in [1,n] ending in digit d */
	int tmp;
	tmp=(n+10-d)/10;
	if(d==0)tmp--;	/* forget 0 */
	return tmp;
	}

ipow(x,y){
/* x^y mod 10 */
	if(y==0) return 1;
	if(y==1) return x;
	return p[x-2][(y-1)%4];
	}