Show Sylvester sequence is the smallest solution with n terms to sum of unit fractions equalling 1 I want to show that a prefix of Sylvester's sequence gives the "smallest" solution to the equation where the sum of n unit fractions equals 1.
$$\sum_{i=1}^{n-1}{\frac{1}{x_i}} + \frac{1}{x_n - 1} = 1$$
Where the terms can be defined in two equivalent ways
$$x_n = \prod_{i=1}^{n-1}{x_i} + 1 = x_{n-1}(x_{n-1} - 1) + 1$$
and our base case is $x_1 = 2$.
For example, with 4 terms
$$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} = 1$$
I define the smallest solution to be the same ordering you may give a vector or tuple, the smaller of two solutions is the one with the first smaller denominator. As examples of this, the below table of fractions (which are not solutions) is sorted from smallest to largest.
$$\begin{array}{|c|c|}
\hline
\frac{1}{2} + \frac{1}{3} + \frac{1}{7} & (2, 3, 7) \\ \hline
\frac{1}{2} + \frac{1}{3} + \frac{1}{8} & (2, 3, 8) \\ \hline
\frac{1}{3} + \frac{1}{7} + \frac{1}{20} & (3, 7, 20) \\ \hline
\frac{1}{3} + \frac{1}{8} + \frac{1}{10} & (3, 8, 10) \\ \hline
\end{array}$$
I can show via induction that the first n terms of Slyvester's sequence always yields a valid solution, but I'm having trouble showing it is the smallest out of all possible solutions with n terms.
What I tried
I began by trying to prove that if you take the smallest $n-1$ term solution and extended it by a term through splitting the $n-1^{th}$ term in two, using the greedy method, you would get the same solution as is given by the first $n$ terms of Slyvester's sequence. While this works out, I can't justify that this method gives the smallest solution with $n$ terms.

Edit
Looking at the above wiki pages, it mentions the first n terms of the slyvester sequence gives the closest approximation to 1 possible with n unit fractions. Is it possible to build a proof based on this?
 A: If I read your question correctly, then it is a different question than what Mathlove answered. You are asking for the following:   

Suppose that $  a_1 \leq a_2 \leq \ldots  \leq a_n $ are integer solutions to$ \sum \frac{1}{a_i} = 1 $. Order them in lexicographic order. Then, the smallest solution is $\{s_i \} = ( x_1, x_2, \ldots x_{n-1}, x_n-1) $.

If so, then you essentially have the proof, but just need to verbalize it out.
Claim 1: $ \sum_{i<n}  \frac{1}{x_i} + \frac{1}{x_n-1} = 1 $.
Proof by induction. Observe that $\frac{1}{x_n -1 } = \frac{1}{ x_n} + \frac{1}{x_n(x_n -1 )} = \frac{1}{x_n} + \frac{1}{x_{n+1} - 1 }$.  
Claim 2: $s_j$ is the smallest integer strictly larger than $ \frac{1}{ 1 - \sum_{i< j} \frac{1}{s_i}}$.
Proof: This is what the greedy algorithm means. It follows directly from Claim 1 since $ 1 - \sum_{i<j} \frac{1}{s_i} = \frac{1}{s_j - 1 }$, so the smallest integer strictly larger than the inverse of that is clearly $s_j$.  
Corollary: Suppose that $\sum \frac{1}{ a_i} = 1 $, and that $\forall i$ satisfying $ 1 \leq i \leq j < n-2$, we have $a_i = s_i$. Then we can conclude that $a_{j+1} \geq s_{j+1}$.
Note: This is what the greedy algorithm means.
Proof: For $j<n-2$, we have $\frac{1}{a_{j+1}} = 1 - \sum_{i<j+1} \frac{1}{s_i}  - \sum_{i>j+1} \frac{1}{a_i} < 1 - \sum_{i<j+1} \frac{1}{s_i}$.
Hence, by claim 2, $s_{j+1} \leq a_{j+1}$. 
Corollary: The smallest lexicographic solution is $\{s_i\}$.
Proof: By claim 1, $\{s_i\}$ works.
By the previous corollary, the first $n-1$ terms have to be equal to $s_i$. Hence, the last term is equal to $s_n$. 

Note: Mathlove is trying to answer:

Suppose that $a_1 \leq a_2 \leq \ldots \leq a_n$ are integers satisfying $ \sum \frac{1}{a_i} < 1 $. Then $\sum\frac{1}{a_i} \leq 1 - \frac{1}{a_{n+1} -1 }$.

I believe this is a correct claim. 
