Lambert's Original Proof that $\pi$ is irrational. I am trying to find Lambert's original proof that $\pi$ is irrational. Wikipedia has a little description but it is quite lacking. Can someone direct me to Lambert's original proof or post his proof here?
Thanks.

Here is the complete proof:
Part I Derivation of continued fraction for $\tan(x)$
Part II Proof of Irrationality of continued fractions
Part III Proof of Irrationality of $\pi$.
Part I
We show 
$$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$
First we treat Lambert's continued fraction expression for the quotient of two power series.
Let $F_0$ and $F_1$ be two power series we may assume without loss of generality that they both have nonzero constant term
and that in fact  both have constant term of $1$.
Define
$$F_1-F_0=b_1 x F_2$$
$$F_2-F_1=b_2 x F_3$$
$$F_3-F_2=b_3 x F_4$$
$$F_{n+1}-F_n =b_{n+1} x F_{n+2}$$
where at each step we make the assumption that $F_n$ has a constant term of $1$. In general we would simply get higher powers of $x$ in the equation.
Define $$G_n=\frac{F_{n+1}}{F_n}$$ then we have by dividing the last equation by $F_{n+1}$,
$$G_n=\frac{1}{1-b_{n+1}xG_{n+1}}$$
therefore
$$\frac{F_1}{F_0}=G_0=\cfrac{1}{1-
\cfrac{b_1 x}{1-\cfrac{b_ 2 x}{1-\cfrac{b_3 x}{1-\cdots}}}}$$
Now we shall apply this to obtain continued fraction expansions for several functions including $\tan x$.
Let us define the particular power series,
\begin{equation*}
\begin{split}
F_n =&1+ \frac{x}{1! (\gamma +n)} +\frac{x^2}{2! (\gamma +n)(\gamma +n+1)} +\frac{x^3}{3! (\gamma +n)(\gamma +n+1)(\gamma +n+2)} \cdots \\
&1+ \sum\limits_{k=1}^{\infty} \frac{x^k}{k! (\gamma +n)\cdots (\gamma +n+k-1)}
\end{split}
\end{equation*}
Then we have
$$F_{n+1}-F_n =-\frac{x}{(\gamma +n)(\gamma +n+1)} F_{n+2}.$$
Thus with $b_n =  -\frac{x}{(\gamma +n)(\gamma +n+1)}$ we get
\begin{equation*}
\begin{split}
\frac{F_1}{F_0}=&\cfrac{1}{1+\cfrac{\frac{x}{(\gamma)(\gamma +1)}}{1+\cfrac{\frac{x}{(\gamma +1)(\gamma +2)}}{1+\cfrac{\frac{x}{(\gamma +2)(\gamma +3)}}{1+\cdots}}}} \\
&=\cfrac{\gamma}{\gamma +\cfrac{x}{(\gamma+1)+\cfrac{x}{(\gamma +2)+\cfrac{x}{(\gamma +3)+\cdots}}}}
\end{split}
\end{equation*}
Now let us set $\gamma=\frac{1}{2}$ and instead of $x$ write $-\frac{x^2}{4}$
Then $$F_1= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)!}=\frac{\sin x}{x}$$
and
$$F_0= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}=\cos x.$$
Putting this altogether we get
$$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$
Part II
In the continued fraction,
$$\cfrac{b_1}{a_1 -
\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}}  $$
assume that
where $1+b_n \leq a_n$ for all $n$, and that we have $1+b_n < a_n$ infinitely often.
Then the fraction is irrational.
Assume that the fraction is rational, say
$$\frac{\lambda_1}{\lambda_0}=\cfrac{b_1}{a_1 -
\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}} $$
where $\lambda_1$ and $\lambda_0$ are positive integers, now since the fraction converges to a number less than one, $\lambda_1 < \lambda_0$
If we set $$\rho_1=\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}} $$
then we have
$$\frac{\lambda_1}{\lambda_0}=\frac{b_1}{a_1 -\rho_1}$$
so$$\rho_1=\frac{a_1 \lambda_1 - b_1 \lambda_0}{\lambda_1} < 1$$
So $\rho_1=\frac{\lambda_2}{\lambda_1}$ where $\lambda_2 < \lambda_1$.
Continuing in this way we obtain a strictly decreasing sequence of positive integers, $\lambda_0 > \lambda_1 > \cdots$, a contradiction.
Part III
This expression leads to the following fundamental result.
$\pi$ is irrational.
Assume that $\pi$ is rational, then $\frac{\pi}{4}$ is also rational.
Let $$\frac{\pi}{4}=\frac{a}{b},$$
and substitute $x=\frac{\pi}{4}$ into Lambert's continued fraction for $\tan x$.
We get
\begin{equation*}
\begin{split}
1=&\cfrac{\frac{a}{b}}{1-
\cfrac{\frac{a^2}{b^2 }}{3-\cfrac{\frac{a^2 }{b^2}}{5-\cfrac{\frac{a^2}{b^2}}{7-\cdots}}}} \\
&=\cfrac{a}{b-
\cfrac{a^2}{3b-\cfrac{a^2}{5b-\cfrac{a^2}{7b-\cdots}}}} \\
\end{split}
\end{equation*}
 Now since eventually $nb > a^2 +1$ we have that this expression is irrational, and this is absurd since it is equal to $1$.
 Therefore $\pi$ is irrational.
