What is the time complexity of the following algorithm

Question

The Question

I am having trouble finding the time complexity of the following algorithm:

void f2(int n) {
    int i;
    if (n == 1) {
        return;
    }
    for (i = 0; i < n; i++) {
        f2(n - 1);
    }
}

My Problem

The solution I came to is O(2^n) but the right answer is stated as O(n!) and I do not understand how that is possible and would be really grateful if somebody could explain it to me.

Answer 1

for (i = 0; i < n; i++) {
        f2(n - 1);
    }

so outer loop runs n times, and recursively calls (n times) a inner loop which runs n-1 times, until you reach 1.

So it runs n! times all right.

Answer 2

Consider below function

void fact(int n) {
   int i;
   if (n == 1) {
      return;
   }
  for(int i=0; i<n; i++) {
    fact(n-1);
  }
}

Assume n=2 ;

the function becomes

void fact(int n) {

    fact(1);
    fact(1);

}

Seems like time complexity is exponential.

Now Assume n=4

void fact(int n) {

    fact(3); 
    // -- this leads to fact(2) fact(2) fact(2) 
    // -- which leads to fact(1) fact(1) fact(1) fact(1) fact(1) fact1)

    fact(3);
    fact(3);
    fact(3);
}

ultimately it ends up with O(n!)