Sample space of possible outcomes for a knockout tournament I would like to confirm if my answer is correct for the following question:

A conventional knock-out tournament begins with $2^n$ competitors and has $n$ rounds. There  are no play-offs for the positions $2,3,..,2^{n-1}$, and the initial table of draws is  specified. Give a concise description of the sample space of all possible outcomes.

My answer is: $$\sum_{i=1}^{n} 2^{2^{i}/2}.$$
My reasoning is as follows: For each round, the number of players is $2^n$. The number of matches for each round is $2^n / 2$. Since there can be only a win or loss, the total number of combinations per round is $2^{2^{n}/2}$. Therefore the total number of combinations of outcomes (the total sample space), is the sum of combinations in each round from $1 \ldots n$.
 A: We have $2^n$ players, so basically there are $\frac{2^n}{2} = 2^{n-1}$ games in the first round. The combination of possible players going to the next round is given by $2^{2^{n-1}}$ because in each game we have two possible winners.
In the second round, we will have $2^{n-1}$ players and $2^{n-2}$ games. Following the same reasoning, the combination of winners is given by $2^{2^{n-2}}$.
And so on, until we reach the final, for which we have $2^1$ possible winners. 
Therefore, the number of possibilities we have is given by the summation of all possible outcomes in different rounds:
$\displaystyle \sum_{i=1}^{n} 2^{2^{n-i}}$.
If we take only the power, we can see that it follows a geometric series form, so we can calculate the sum of all powers $= 2^n - 1$
Therefore, the final number of possibilities is $2^{2^n - 1}$.
The explanation of Tony K is much simpler, however I couldn't figure out why the number of games was $2^n - 1$ until I developed this reasoning.
Hope it helps!
A: You don't need to sum any complicated series here. Simply observe that each match eliminates one competitor, therefore there are $2^n-1$ matches, each of which has two possible outcomes.
A: A nice question and answers! In the original question, the initial draw is specified. Here is an answer if the initial draw is not specified and we want the sample space to also account for different possible draws (and then outcomes for a given draw). In the first round there are
$$
\frac{1}{(2^{n-1})!}{2^{n}\choose 2,\dots,2}=\frac{(2^n)!}{(2^{n-1})!2^{2^{n-1}}}
$$
ways to pair the players and $2^{2^{n-1}}$ possible first round results, producing $2^{n-1}$ first round winners. Similarly, if the draw is not specified initially we have 
$$
\frac{1}{(2^{n-2})!}{2^{n-1}\choose 2,\dots,2}=\frac{(2^{n-1})!}{(2^{n-2})!2^{2^{n-2}}}
$$
ways to pair the first round winners for the second round, and $2^{2^{n-2}}$ given a fixed second round set of parings. Continuing in this way until the final, the total sample space if we account for different possible draws and then outcomes for a given draw is
$$
\prod_{i=1}^{n}\left(\frac{(2^{n-(i-1)})!}{(2^{n-i})!2^{2^{n-i}}}\times 2^{2^{n-i}} \right)=(2^n)!
$$
which is the number of permutations of the competitors. One way to produce a bijection with this sample space and permutations of the competitors is that for a given draw and outcome for the draw, we may rank the tournament winner as first, the other finalist as second, the semifinalist who lost to the winner as third, the semifinalist who lost to the person ranked second as fourth, the quarterfinalist who lost to the winner as fifth, and so on.
