How prove this $b_{n}>b_{n+1}$, if $a_{1}=1,a_{n+1}=a_{n}+e^{-a_{n}},b_{n}=a_{n}-\ln{n}$ Question:

let sequence 
  $$a_{1}=1,a_{n+1}=a_{n}+e^{-a_{n}}$$
  let $$b_{n}=a_{n}-\ln{n},n\in Z$$
show that:
  $$ b_{n}>b_{n+1}$$
  where $$\ln{n}=\log{n}$$

my idea: since $$b_{n}=a_{n}-\ln{n}$$
then we have
$$b_{n+1}=b_{n}+\dfrac{1}{n}e^{-b_{n}}-\ln{\dfrac{n+1}{n}}$$
then we only prove
$$e^{-b_{n}}\le\ln{\left(1+\dfrac{1}{n}\right)^n}$$
and we only prove
$$b_{n}\ge -\ln{\ln{\left(1+\dfrac{1}{n}\right)^n}}?$$
then I can't.
 A: Here is the beginning of a potentially helpful idea which I unfortunately cannot flesh out at the moment. I will come back in a while and try to do so.
Consider the functions 
$$
  f(x)
= x + e^{-x}, \quad 
  g(x)
= \ln(1 + e^x).
$$
Then $a_n = f \circ f \circ \cdots \circ f(0) = f^n(0)$ and $g^n(0) = \ln(n+1)$, so the condition
$$
     e^{-a_n} 
\leq \ln\left(\frac{n+1}{n} \right)
=    \ln(n+1) - \ln(n)
$$
is equivalent to 
$$
     f^{n+1}(0) - f^n(0)
\leq g^n(0) - g^{n-1}(0).
$$
This inequality could be established by establishing 
$$
     g^n(x) - g^{n-1}(x) -( f^{n+1}(x) - f^n(x))
\geq 0 \quad \quad \quad (*)
$$
for all $x \geq 0$ and simply taking $x=0$. Since this expression converges to $0$ as $x \to \infty$, it is sufficient to show that the derivative is negative. I believe this can be done by induction on $n$: by the chain rule
$$
  \frac{d}{dx} f^{n+1}(x)
= (f^n)'(f(x))f'(x)
= (1 - e^{-x})(f^n)'(f(x))
$$
and therefore 
$$
  \frac{d}{dx} (f^{n+1}(x) - f^n(x))
= (1-e^{-x})((f^n)' - (f^{n-1})')(f(x)).
$$
Similarly,
$$
  \frac{d}{dx}(g^n(x) - g^{n-1}(x))
= \frac{e^x}{1+e^x}((g^{n-1})'-(g^{n-2})')(g(x)).
$$
I'm not sure where to go from here yet.
A: Proof:
Fact 1: It holds that $a_n > - \ln \ln \frac{n+1}{n}$. (The proof is given at the end.)
From Fact 1, we have
\begin{align}
b_n - b_{n+1} &= a_n - \ln n - (a_{n+1} - \ln (n+1))\\
&= a_n - \ln n - a_n - \mathrm{e}^{-a_n} + \ln (n+1)\\
&= \ln \frac{n+1}{n} - \mathrm{e}^{-a_n}\\
&> 0.
\end{align}
(Q. E. D.)
$\phantom{2}$
Proof of Fact 1: We use the mathematical induction.
For $n = 1$, we need to prove that $a_1 > - \ln \ln 2$. It is true.
Assume that the inequality is true for $n = k$ ($k\ge 1$), i.e., $a_k > -\ln \ln \frac{k+1}{k}$.
Let us prove that the inequality is also true for $n = k + 1$, i.e., $a_{k+1} > -\ln \ln \frac{k+2}{k+1}$.
Since $x\mapsto x + \mathrm{e}^{-x}$ is strictly increasing on $(0, \infty)$, we have
\begin{align}
a_{k+1} &= a_k + \mathrm{e}^{-a_k}\\
& > -\ln \ln \frac{k+1}{k} + \mathrm{e}^{-(-\ln \ln \frac{k+1}{k})}\\
&= -\ln \ln \frac{k+1}{k} + \ln \frac{k+1}{k}.
\end{align}
Thus, it suffices to prove that
$$-\ln \ln \frac{k+1}{k} + \ln \frac{k+1}{k} > -\ln \ln \frac{k+2}{k+1}$$
or
$$\frac{k+1}{k}\ln \frac{k+2}{k+1} > \ln \frac{k+1}{k}$$
or
$$\frac{1}{k}\ln \frac{k+2}{k+1} > \ln \frac{k+1}{k} - \ln \frac{k+2}{k+1} = \ln\left(1 + \frac{1}{k^2+2k}\right).$$
Since $\ln (1+x) < x$ for $x > 0$, it suffices to prove that
$$\frac{1}{k}\ln \frac{k+2}{k+1} > \frac{1}{k^2+2k}$$
or
$$\ln\left(1 + \frac{1}{k+1}\right) > \frac{1}{k+2} = \frac{\frac{1}{1+k}}{1 + \frac{1}{k+1}}$$
which follows from $\frac{x}{1+x} < \ln (1+x)$ for $x > 0$. We are done.
