Explanation for crashing at n = 16 - recursion

Question

I am having a trouble in completing a homework of mine in C.The task is given an integer n print all the binary numbers with the length n that do not have 2 consecutive zeros int them and that includes leading zeros, note that at least one of the functions has to be recursive. Here is an example if n is 4 then the binary number 10 is treated 0010 and therefore wont be printed because it has 2 leading zeros. My problem is that my code crashes if n = 16 and I do not know why even though I have done a lot of debugging. Here is my code, Thanks for any help here.
void binaries_n_digits_no_00(int n)
{
if(n < 0)
{
return;
}
print_binaries_n_digits_no_00(0,n);
}
void print_binaries_n_digits_no_00(long int current_binary_index,int n)
{
int num_of_leading_zeros;
if((current_binary_index > (power(2,n) - 1)) || n == 0)
{
return;
}
num_of_leading_zeros = n - binary_num_length(current_binary_index);
if((binary_not_contain_00(current_binary_index) == 1) &&
(num_of_leading_zeros == 1 || num_of_leading_zeros == 0)){
if(current_binary_index == 0 && n == 1)
{
printf("\n0");
}
else if(num_of_leading_zeros == 1)
{
if(current_binary_index != 0)
{
printf("\n0");
print_binary(current_binary_index);
}
}
else{
printf("\n");
print_binary(current_binary_index);
}
}
print_binaries_n_digits_no_00(current_binary_index+1,n);
}
int binary_not_contain_00(long int num)
{
if(num/2 == 0)
{
return 1;
}
if(((num%2) == 0) && (((num/2) % 2) == 0))
{
return 0;
}
return binary_not_contain_00(num/2);
}
void print_binary(long int num)
{
if(num > 1)
{
print_binary(num/2);
}
printf("%d",num%2);
}
int binary_num_length(long int num)
{
if(num <= 1)
{
return 1;
}
else{
return (1 + binary_num_length(num/2));
}
}
long int power(int m, int n)
{
if(n == 0)
{
return 1;
}
return m * power(m,n-1);
}