This is my solution:

#include <stdio.h>

long double fact( long double n )
{
    if (n < 2)
        return 1;
    else
        return n * fact(n - 1);
}

int main( void )
{
    long double n, nfac, ans = 0, i;

    scanf("%Lf", &n);

    if (n < 2)
    {
        printf("0");
        return 0;
    }

    nfac = fact(n);

    for (i = 2; i < n; i ++)
        ans += nfac / fact(n - i);

    ans += nfac;

    printf("%.0Lf", ans);

    return 0;
}

WTF?

Check your answers to maxtests:

21 => 138879579704209680000
20 => 6613313319248079980

PS: Idea of solution is correct, but you should store answer in suitable type ("long double" is not enough).

Hmm, I actually have correct answers for n = 20 and 21.
And "long double" is extremely big type, for example:
100 =>
253686955560127297296368144135170000933446331847165085913916476338770833469945019420289324705338829425834316771115322447713278811189027647015804045741541818368
=))
What's wrong then?

"long double" can store numbers about 10^{4000} but only first 19 digits are calculated right.

Btw, I think, you use some else compiler, not Intel-C++, am I right?

Yep, you're right: I use GNU GCC.
Ok, I will try to use "long long" type.

you are so stupid max n is 21 not 100