What is the least significant non-zero digit in the decimal expansion of n!?

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];
        }