$|\sin(\sin( \cdots \sin(x)\cdots))|$ ($N$ times) is always $\leq|\sin( \cdots \sin(1)\cdots)|$ ($N-1$ times) 
Is this inequality always true?
  $$
\bigl\lvert\,\underbrace{\sin(\sin(\cdots \sin}_{N\text{ times}}(x)\cdots))\bigr\rvert\le\bigl\lvert\,\underbrace{\sin(\sin( \cdots \sin}_{N-1\text{ times}}(1)\cdots))\bigr\rvert
$$

I think so, for example for $N=2$ we have:
$|\sin(x)|\leq 1$ therefore $|\sin(\sin(x))|\leq |\sin(1)|$
 A: Sorry, but I remember you that the $\sin$ function is strictly increasing only in $Z_1=\displaystyle\bigcup_{k\in\mathbb{Z}}\left[-\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi\right]$ and therefore:
\begin{equation}
\forall x\in Z_1,\,\sin(\sin x)\leq\sin(1),\,1>\sin 1>0;
\end{equation}
while in $Z_2=\displaystyle\bigcup_{k\in\mathbb{Z}}\left[\frac{\pi}{2}+2k\pi,\frac{3\pi}{2}+2k\pi\right]=Z_1+\pi$ the $\sin$ function is strictly decreasing and therefore:
\begin{gather}
\forall x\in Z_2,\exists y\in Z_1:x=y+\pi\Rightarrow\\
\Rightarrow\sin(\sin(x))=\sin(\sin(y+\pi))=\sin(-\sin(y))=-\sin(\sin(y))\geq-\sin(1).
\end{gather}
From all this:
\begin{equation}
\forall x\in\mathbb{R},\,|\sin(\sin(x))|\leq|\sin1|,
\end{equation}
and by induction you can prove the claim.
Is it all clear?
A: First fact to use:

For $x\ge0$, $\sin(x)\le x$

Second fact to use:

$1<\pi/2$

Third fact to use:

The sine function is increasing in the interval $(0,\pi/2)$

Also, it's not restrictive to assume $x\ge0$, because $\sin(-x)=-\sin(x)$ and the minus sign is “eaten” by the absolute value.
For $x\ge0$, we have $0\le \sin(x)\le 1$, which implies the simplest case of your statement, namely that
$$
\sin(\sin(x))\le\sin(1)
$$
because we are in the interval $(0,\pi/2)$ where the sine is increasing.
For the general case, denote by $\sin^{[n]}$ the “$n$-fold iteration”:
$$
\sin^{[1]}(x)=\sin(x),\qquad
\sin^{[n+1]}(x)=\sin(\sin^{[n]}(x))
$$
so the statement above can be written
$$
\sin^{[2]}(x)\le\sin^{[1]}(1)
$$
which we can now use for a proof by induction of your general statement, namely that
$$
\sin^{[n+1]}(x)\le\sin^{[n]}(1)
$$
for $n\ge1$.
First a side step:
$$
0\le\sin^{[n]}(x)\le 1
$$
for $n\ge1$. Indeed, this is true for $n=1$. Suppose it is for some $n\ge1$; then
$$
\sin^{[n+1]}(x)=\sin(\sin^{[n]}(x))\le \sin^{[n]}(x)
$$
from the first fact. By the induction hypothesis, $\sin^{[n]}(x)\le 1$.
Now, suppose we know that, for some $n\ge1$,
$$
\sin^{[n+1]}(x)\le\sin^{[n]}(1)
$$
Then
$$
\sin^{[n+2]}(x)=\sin(\sin^{[n+1]}(x))
\le
\sin(\sin^{[n]}(1)
=
\sin^{[n+1]}(1)
$$
where we use that $\sin^{[n]}(1)\le1$, as proved in the side step.
