Proof of the extrinsic to intrinsic rotation transform Wikipedia states that:

Any extrinsic rotation is equivalent to an intrinsic rotation by the
  same angles but with inverted order of elemental rotations, and
  vice-versa. For instance, the intrinsic rotations $x-y’-z''$ by angles $\alpha, \beta,\gamma$ are equivalent to the extrinsic rotations $z-y-x$ by angles $\gamma, \beta, \alpha$. 

Is there a simple proof why this is so?
 A: The statement is valid not only for rotations. According to relativity,
there is no preferred frame of reference / coordinate system.
Therefore, in kinematics, all (times and) positions and motions are relative. (Dynamics - with forces - is another matter).
Consider a two-dimensional example, as has been found in a book about
Computer Graphics (to be precise: J.D. Foley, A. van Dam, Fundamentals
of Interactive Computer Graphics, 1982). There are two frames of reference,
one attached to the observer (world), one attached to an object (chair):

With a transformation of coordinates, the only thing that is important is the
relative position of the object with respect ot the observer. This means that
the end result of a coordinate transformation can be achieved in at least two ways.
As is displayed in the example:

Extrinsic. 
Rotate $\,R\,$ the chair in the world coordinate system over an angle of $45^o$
and then Translate $\,T\,$ it over a distance $(4,10)$. Thus resulting in a transformation $\,TR$ .

Intrinsic. 
Translate $\,T^{-1}\,$ the observer in the chair coordinate system over a distance
$(-4,-10)$ and then rotate $\,R^{-1}\,$ the world coordinate system over an angle $-45^o$.
Thus resulting in a transformation $\,R^{-1}T^{-1}$ .

With this simple example, we can see immediately that the transformations
are the inverse of each other: $\,R^{-1}T^{-1}= (TR)^{-1}$ .
Hope you get the idea. 
Generalizing this to three dimensions is expected to be a matter of filling in the
(somewhat more involved) technicalities.

Update. Hmm, "not quite a proof". Then perhaps this.
Let the coordinate system of the object be called $O$ and the world coordinate
system be called $W$. Both coordinate systems are coincident in the beginning.
The first step is to apply a transformation $R$ to $O$  
(just as in the example, but in general now).The second step is to make $W$
coincident again with $O$, which is done by applying the same transformation
$R$ to $W$ as has been done in the first step with $O$.Then effectively nothing
has changed and we have the original configuration again: the product of step (1)
and step (2) is the identity.It is thus obvious that
the first step could also have been accomplished by applying the inverse $R^{-1}$
transformation to $W$ instead of $O$.Common
properties of inverse operations like $(AB)^{-1}= B^{-1}A^{-1}$ are assumed throughout to be well known. This completes the proof.
A: I googled on the Internet, but also found not much material on this topic.
The most relevant references are from Wikipedia as @Apoo suggested, and a blog the proof is at bottom of the page.
Though the arguments are quite complete, neither of them could convince me. In order to prove the statement, I have to first introduce changing a transformation between coordinate systems. There is a derivation from blender stackexchange, and I excerpt the equation as follows:
$$
T_{\text {world}}=S_{\text {world}} \times T_{s} \times S_{\text {world}}^{-1}, \tag{1}
$$
where $T_{\text {world}}$ is the transformation matrix in world coordinate system,
$S_{\text {world}}$ is the world matrix of the referenced object $s$,  $T_{s}$ is the local transformation matrix based on $s
$.
Assume that an intrinsic Euler angle can be represented as a product of three rotation matrices
$$
R = Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha).
$$
The goal is to prove, there exists an extrinsic rotation sequence, s.t.
$$
Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha) =
X(\alpha)Y(\beta)Z(\gamma).
$$
Consider $Y^\prime(\beta)$ as a rotation matrix around $y^\prime$ of angle $\beta$ relative to $x^\prime - y^\prime - z^\prime$ coordinate system, we can obtain the corresponding rotation matrix in $x-y-z$ coordinate system using conversion eq. $(1)$:
$$
Y^\prime = X Y X^{-1},
$$
it follows that
$$
Y =  X^{-1} Y^\prime X . \tag{2}
$$
Through a similar argument, we can also get the world rotation matrix of angle $\gamma$ around $z^{\prime \prime}$ in $x-y-z$ coordinate system:
$$
Z^{\prime \prime} = (Y^\prime X) Z (Y^\prime X)^{-1},
$$
and it follows that
$$
Z = (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X). \tag{3}
$$
Multiplying $X$ by eq. $(2)$ and eq. $(3)$, we obtain
$$
\begin{align}
X Y Z &= X X^{-1} Y^\prime X (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X) \\
      &= Z^{\prime \prime} Y^{\prime} X ,
\end{align}
$$
which completes the proof.
