Why second loop of sieve of Eratosthenes starting from square of current prime number?

vector<bool>vc(100006,1);
void seive(int n)
{
    vc[0]=vc[1]=0;
    int i,j;
    for(i=2;i*i<=n;i++)
    {
        if(vc[i]==1)
        {
            for(j=i*i;j<=n;j=j+i)
            {
                vc[j]=0;
            }
        }
    }

}

My question is why the second loop starting from square of i. How the multiples before the square of current prime(i) getting marked ?Actually how are we so sure of it?