What went astray in this code — based on Proofwiki — but I tried to edit?

Question

I'm trying to fix the code for my question from http://www.proofwiki.org/wiki/Euler%27s_Number_is_Irrational I can't see what went astray. I didn't just copy and paste this code
— I shortened it. Furthermore, can you please expose how you determined the problem?

Multiplying both sides by $n!$ $\color{red}{\text{Why — eliminate the denominators? Please edit this if wrong}}$:

$\displaystyle \frac m n n! = n! \sum_{i \mathop = 0}^\infty \frac 1 {i!} = \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac {n!}{n!}}\right) + \left({\frac{n!}{\left({n + 1}\right)!} + \frac{n!}{\left({n + 2}\right)!} + \frac{n!}{\left({n + 3}\right)!} + \cdots}\right)$

$ {{eqn | l=m \left({n - 1}\right)! - \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac{n!}{n!} } =\frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right) \left({n + 3}\right)} + \cdots |=\frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right) \left({n + 1}\right)} + \cdots =\sum_{i \mathop = 0}^\infty \left ({\frac 1 {n+1} }\right)^{\left({i + 1}\right)} r=\frac{\frac 1 {n+1} } {1 - \frac 1 {n+1} } = \frac 1 n < 1\) from [[Sum of Infinite Geometric Progression]] \( | c= }} $

This is what I want to work out —

Answer 1

The main problem is the second \left( after the eqn is unbalanced. If you remove that, most of the codes will be rendered properly as illustrated below:

$$ {{eqn | l=m \left({n - 1}\right)! - {\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac{n!}{n!} } =\frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right) \left({n + 3}\right)} + \cdots |=\frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right) \left({n + 1}\right)} + \cdots =\sum_{i \mathop = 0}^\infty \left ({\frac 1 {n+1} }\right)^{\left({i + 1}\right)} r=\frac{\frac 1 {n+1} } {1 - \frac 1 {n+1} } = \frac 1 n < 1\) from [[Sum of Infinite Geometric Progression]] \( | c= }}$$