Finding the limit of recursive sequence Assuming that the solution of $e^{-x}=x$ is $c\in (0,1)$

And give the following sequence

$$a_{n+1}=e^{-a_n}$$

$$a_1=1$$

How can i prove that the sequence converge and that the limit is  $$\lim_{n \to \infty}a_n=c$$

I tried to define a new sequence $$b_n=a_n-c$$ and to prove it using the ratio test but without success
$f(x) = \mathrm{e}^{-x}$ is a contraction in the relevant range $(0, 1]$, thus $x_{n + 1} = f(x_n)$ converges if started inside that range.
Exponential function is continuous,then

$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}e^{-a_n}=e^{-\lim_{n\to\infty}a_n}.
$$

So the limit of this sequence is the solution of $x=e^{-x}$ which is $c$.

To see the existence, first note by induction that $1\geq a_n\geq 0$. Now we claim that subsequence with odd terms decreasing and subsequence with even terms increasing. For $k=1$, $
a_{2k-1}\geq a_{2k+1}
$ is immediate and to see $a_{2k}\leq a_{2k+2}$ note that $1\geq a_n$ and $e$ is monotone. Thus, $
a_2=e^{-1}\leq a_4=e^{-a_3}.
$

Suppose now for some $k$, $a_{2k-1}\geq a_{2k+1}$ and $a_{2k}\leq a_{2k+2}$. Then

$$
e^{-a_{2k}}\geq e^{-a_{2k+2}}\implies a_{2k+1}\geq a_{2k+3},\\
e^{-a_{2k+1}}\leq e^{-a_{2k+3}}\implies a_{2k+2}\leq a_{2k+4}.
$$

This tells us that both $\lim_{k\to\infty}a_{2_k}=l_0$ and $\lim_{k\to\infty}a_{2k-1}=l_1$ exist by monotone convergence theorem. Now we claim that every odd term is bigger than every even term. Trivially for $k=1$, this claim holds. Suppose now that it is true for some $k$, $a_{2k-1}\geq a_{2k}$. Then 

$$
a_{2k+1}=e^{-a_{2k}}\geq e^{-a_{2k-1}}=a_{2k}\implies e^{-a_2k}=a_{2k+1}\geq e^{a_{2k+1}}=a_{2k+2}
$$
which shows that the claim is true for $k+1$ and induction is completed. So we show that $a_{2k}\leq a_{2k+2}\leq a_{2k+1}\leq a_{2k-1}$. Note also that $l_0$ is supremum and $l_1$ is infimum for corresponding sequences. Then $l_1\geq l_0$.

By induction one can show $a_{2k+1}=a_{2k}^{1/e}\geq a_{2k+1}^{1/e}=a_{2k+2}$. Taking limit on both sides  gives us $l_0^{1/e}\geq l_1^{1/e}$ and $l_0\geq l_1$. Thus $l_0=l_1=l$ and $a_k\to l$.