*spoiler alert*  solution and answer follow below

It’s straightforward to extend the logic above to the general case when we have \(2^N\) teams at the start.

Let \(f(2^N) \) be the function giving us the number of ways in which \( 2^N \) teams can be ranked. 

Now we know that \(f(2^{1}) = 1 \) and \( f(2^{2}) = 3 \). 

We can use the logic used in the section above to determine \( f(2^{N}) \) knowing \( f(2^{N-1}) \). This is because the \( 2^{N}\) case is just two separate \( 2^{N-1} \) team single eliminations with the winners meeting in the finals to definitely determine the top team. 

\( f(2^{N-1}) \) is then determined as the number of ways to choose \( 2^{N-1} \) slots from \( 2^{N} – 1\) slots and multiplying it by the number of ways each of the \( 2^{N} – 1\) teams can be ranked.  

\begin{align*}
f(2^{N}) &= {2^{N} – 1 \choose 2^{N-1}} \cdot f(2^{N-1}) \cdot f(2^{N-1}) \qquad &&(N \geq 1) \\
&= {2^{N} – 1 \choose 2^{N-1}} \ f(2^{N-1})^2 \qquad &&(N \geq 1) \\
f(2^{1}) &= 1
\end{align*}

This recursive relation is all we need to calculate the number of possible rankings for any \(2^N\) teams completing the single elimination bracket.

Determining a closed form formula for \(f(2^N) \) requires a bit of algebraic manipulation and series summation, the outline of which I provide below:

Using the recursive relationship above we can see/show:

\begin{align*}
&f(2^2) \ = \ {3 \choose 2} \ 1^2 \ = \ {3 \choose 2} \\
&f(2^3) \ = \ {7 \choose 4} \ {3 \choose 2}^2 \ = \ \frac{7!}{4\cdot 2^2} \\
&f(2^4) \ = \ {15 \choose 8} \ \left ( \frac{7!}{4\cdot 2^2} \right )^2 \ = \ \frac{15!}{8\cdot 4^2 \cdot 2^4} \\
. \\
. \\
. \\
& \textit{and in general } \\ \\
&f(2^N) \ = \ \frac{(2^N – 1)!}{(2^{N-1})^1 \cdot (2^{N-2})^2 \cdot (2^{N-3})^4 \ .\ . \ . \ 2^{N-2}} \\ \\
& \textit {or in product notation } \\ \\
&f(2^N) \ = \ \frac{(2^N – 1)!}{\prod_{i=1}^{N-1} (2^{N-i})^{2^{i-1}}}
\end{align*}

The denominator has \(N-1\) exponent terms of 2. The final exponent of 2  can be summed to \(S\) as follows:

\begin{align*}
S &= (N-1)1 \ + \ (N-2)2 \ + \ (N-3)4 \ + \ . \ . \ . \ \ + \ (2)2^{N-3} \ + \ (1))2^{N-2} \\
&= (N-1)2^0 \ + \ (N-2)2^1 \ + \ (N-3)2^2 \ + \ . \ . \ . \ \ + \ (N-(N-2))2^{N-3} \ + \ (N-(N-1))2^{N-2} \\
&=\left [ 2^0N + 2^1N +2^2N + \ … \ +2^{N-3}N + 2^{N-2}N\right ] – \left [ 1\cdot 2^0 + 2\cdot 2^1 + 3\cdot 2^2 + \ … \ + (N-1)2^{N-2} \right ] \\
&= \left [S_a \right ] \qquad – \qquad \left [S_b \right ]
\end{align*}

The first term \(S_a\) is a geometric series of \(N-1\) terms whose sum is:

\[
S_a = N(2^{N-1} – 1)
\]

The second term \(S_b\) is an arithmetico-geomteric sequence whose sum can be calculated as follows:

\begin{align*}
S_b &= 2S_b – S_b \\
&= (N-1)2^{N-1} – \left ( 2^0 + 2^1 +2^2 + \ … \ + 2^{N-2} \right ) \\
&= (N-1)2^{N-1} – (2^{N-1} – 1) \\
&= N2^{N-1} – 2^N + 1
\end{align*}

Our required S is then: 

\begin{align*}
S &= S_a – S_b \\
&= N(2^{N-1} – 1) – N2^{N-1} – 2^N + 1 \\
&= 2^N – N – 1
\end{align*}

Our final required function \(f(2^N) \) can now be written as:

\begin{align*}
f(2^N) &= \frac{(2^N – 1)!}{2^{(2^N – N – 1)}} \\
&= \frac{2^N(2^N – 1)!}{2^{(2^N – 1)}} \\
&= \mathbf{{\color{Red} {\frac{(2^N)!}{2^{(2^N – 1)}}}}}
\end{align*}

which is the required expression.

Notice that this expression is equivalent to: 

\[
\frac{(\textit{total no. of teams})!}{2^{\textit{total number of games}}}
\]

This is not a coincidence. If you start with the total number of ranking permutations i.e. \( (2^N)!\),  and take any feasible match, say A vs B, before the game, there are an equal number of permutations with A ranked higher than B, as B higher than A. But the outcome of the game reduces the feasible ranking permutations by 2. This is true of every game, leading to the answer. This is a straightforward way to derive the final expression.

Applying this to the March Madness round of 64 with N = 6, the number of possible rankings is \( \approx 1.37 \ \times  \ 10^{70} \) which is in the realm of the total number of atoms in the observable universe!