Prove $\frac{c_n(a_1,\ldots,a_n)}{c_{n-1}(a_2,\ldots,a_n)}=a_1 + \frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_{n-1}+\frac{1}{a_n}}}}$ For $n>0$ and $a_1,...,a_n \in K$ let $c_n(a_1,...,a_n)$ be the determinant of the matrix
$$
  \begin{pmatrix}
  a_1 & 1 & 0 & \cdots & 0 \\
  -1 & a_2 & \ddots & \ddots & \vdots \\
  0 & \ddots & \ddots & \ddots & 0 \\
  \vdots & \ddots & \ddots & \ddots & 1 \\
  0 & \cdots & 0 & -1 & a_n
  \end{pmatrix}
$$
Show for $n \ge 2$ that following holds:
$$
  \frac{c_n(a_1,...,a_n)}{c_{n-1}(a_2,...,a_n)}
  = 
  a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_{n-1}+\frac{1}{a_n}}}}
$$
I want to show it using induction over n but I already fail at the initial step:
For $n = 2$ I have to show:
$$
  \frac{c_2(a_1,a_2)}{c_{2-1}(a_2)}
  = 
  a_1 + \frac{1}{a_2}
$$
which is
$$
  \frac{a_1a_2 + 1}{a_2} \neq a_1 + \frac{1}{a_2}
$$
My also have no idea for the induction step where I have to show:
$$
  \frac{c_{n+1}(a_1,...,a_{n+1})}{c_n(a_2,...,a_{n+1})}
  = 
  a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}}
$$
So first I calculate $c_{n+1}(...)$ by developing after the first column which gives me:
$$
  a_1 \cdot \det
  \begin{pmatrix}
  a_2 & 1 & & & & \\
  -1 & a_3 & 1 & & & \\
  & -1 & \ddots & & & \\
  & & & \ddots & & \\
  & & & & & 1 \\
  & & & & -1 & a_{n+1}
  \end{pmatrix}
  +
  \det
  \begin{pmatrix}
  1 & 0 & & & & \\
  -1 & a_3 & 1 & & & \\
  & -1 & a_4 & & & \\
  & & & \ddots & & \\
  & & & & & 0 \\
  & & & & 0 & a_{n+1}
  \end{pmatrix}
$$
which after developing after the first row gives me
$$
  a_1 \cdot c_n(a_2,...,a_{n+1}) + c_{n-2}(a_3,...,a_{n+1}).
$$
Analog for $c_n(...)$:
$$
  c_n(...) = a_2 c_{n-1}(...) + c_{n-3}(...)
$$
So I have
$$
  \frac{c_{n+1}(...)}{c_n(...)} = \frac{a_1c_n(...) + c_{n-2}(...)}{a_2c_{n-1}(...) + c_{n-3}(...)}
$$
written in another way it is
$$
  \frac{a_1c_n(...)}{c_n(...)} + \frac{c_{n-2}}{a_2c_{n-1}(...) + c_{n-3}(...)}
$$
I write it as
$$
  a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{c_{n-3}(...)}{c_{n-2}(...)}}
$$
and then
$$
  a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{1}{\frac{c_{n-2}(...)}{c_{n-3}(...)}}}
$$
so
$$
  a_1 + \frac{1}{a_2 \cdot \left( a_3 + \cfrac{1}{a_4 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right) + \frac{1}{\left( a_4 + \cfrac{1}{a_5 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right)}}
$$
I dont know how to preced from here, any help or simpler solutions are appreciated!


but I already fail at the initial step

Not really. You have
$$\frac{c_2(a_1,a_2)}{c_1(a_2)} = \frac{a_1a_2+1}{a_2} = \frac{a_1a_2}{a_2} + \frac{1}{a_2} = a_1 + \frac{1}{a_2}.$$
In your induction step, you also have the necessary ingredients,
$$c_{n+1}(a_1,\dotsc,a_{n+1}) = a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1}),$$
you just have made a dimension error and thought it was $c_{n-2}(a_3,\dotsc,a_{n+1})$. The dimension of that matrix is $\bigl((n+1)-2\bigr)^2$, not $(n-2)^2$.
And then you rename, calling $b_k = a_{k+1}$ for $k = 1,\dotsc, n$, to get
$$\begin{align}
\frac{c_{n+1}(a_1,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})} &= \frac{a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\
&= a_1 + \frac{c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\
&= a_1 + \frac{1}{\frac{c_n(b_1,\dotsc,b_n)}{c_{n-1}(b_2,\dotsc,b_n)}}\\
&= a_1 + \cfrac{1}{b_1 + \cfrac{1}{b_2 + \cfrac{1}{\ddots + \cfrac{1}{b_{n-1}+ \frac{1}{b_n}}}}}\\
&= a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots + \cfrac{1}{a_{n}+ \frac{1}{a_{n+1}}}}}}
\end{align}$$
using the induction hypothesis on the sequence $b_1,\dotsc,b_n$.
