The most conceptual explanation uses composition of species.
Recall that your Stirling number of the first kind counts collections of $k$ cycles that together contain $n$ points.
Now, you first determine the exponential generating function for cycles:
There are $(n-1)!$ cycles of length $n$, for example, because you regard the list of elements coming after 1 in the cycle.
So, you get $Cycles(z)=\sum_n \frac{(n-1)!}{n!}z^n= \sum \frac{z^n}{n}= \log(\frac 1 {1-z})$.
Then, you determine the exponential generating function for sets of size $k$.
There is only one possibility to do so and it has size $k$:
$Sets_k(z)=\frac{z^k}{k!}$
And, now the miracle occurs:
You want to count $k$-sets of cycles and you find the exponential generating function by calculating $Sets_k$ of $Cycles$ with composition of functions.
So, you get $Sets_k(Cycles(z))= \frac{(\log(\frac 1 {1-z}))^k}{k!}$.
You can read more on this on the corresponding Wikipedia page
or you can read much more on this in the book Combinatorial Species and Tree-Like Structures by F. Bergeron, Gilbert Labelle, Pierre LeRoux.
"Concrete Mathematics (what else?) - Eulerian Numbers" - says:
"Second-order Eulerian numbers are important chiefly because of their connection with Stirling numbers"
Eq. (6.43) therein gives
$$
\left\{ \matrix{ x \cr x - n \cr} \right\}
= \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{
n \cr
k \cr} \right\rangle } \right\rangle } \binom{x+n-1-k}{2n}
\quad \left| \matrix{
\;0 \le n \in Z \hfill \cr
\;x \in C \hfill \cr} \right.
$$
which easily reduce to yours, and can be extended to define Stirling No. of 2nd kind, of the indicated form,
to complex values of $x$.
Interestingly, also given is a twin one for the Stirling No. of 1st kind
$$
\left[ \matrix{ x \cr x - n \cr} \right]
= \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{
n \cr k \cr} \right\rangle } \right\rangle } \binom{x+k}{2n}
\quad \left| \matrix{
\;0 \le n \in Z \hfill \cr
\;x \in C \hfill \cr} \right.
$$
Best Answer
Here is another proof for the curious. We seek to verify that (with $n\ge 1$, $n=0$ holds by inspection)
$$\sum_{k=0}^n \left\langle n \atop k \right\rangle x^{n-k} = (1-x)^n \sum_{k=0}^n {n\brace k} k! \left(\frac{x}{1-x}\right)^k.$$
We get using standard EGFs for the RHS
$$n! [z^n] (1-x)^n \sum_{k=0}^n \frac{(\exp(z)-1)^k}{k!} k! \left(\frac{x}{1-x}\right)^k \\ = n! [z^n] (1-x)^n \sum_{k=0}^n (\exp(z)-1)^k \left(\frac{x}{1-x}\right)^k.$$
Now because $\exp(z)-1 = z+\cdots$ we have $(\exp(z)-1)^k = z^k + \cdots$ so when $k\gt n$ there is no contribution to the coefficient extractor and we get
$$n! [z^n] (1-x)^n \sum_{k\ge 0} (\exp(z)-1)^k \left(\frac{x}{1-x}\right)^k \\ = n! [z^n] (1-x)^n \frac{1}{1-(\exp(z)-1)x/(1-x)} \\ = n! [z^n] (1-x)^n \frac{1-x}{1-x-(\exp(z)-1)x} \\ = n! [z^n] (1-x)^n \frac{1-x}{1-x \exp(z)} \\ = n! [z^n] \frac{1-x}{1-x\exp(z(1-x))}.$$
On the other hand we have for the LHS by the mixed GF of the Eulerian numbers
$$n! [z^n] \sum_{k=0}^n x^{n-k} [w^k] \frac{w-1}{w-\exp((w-1)z)}$$
Now we have $\left\langle n \atop k \right\rangle = 0$ when $k\ge n$ so this is
$$n! [z^n] x^n \sum_{k\ge 0} x^{-k} [w^k] \frac{w-1}{w-\exp((w-1)z)} \\ = n! [z^n] x^n \frac{1/x-1}{1/x-\exp((1/x-1)z)} \\ = n! [z^n] x^n \frac{1-x}{1-x\exp((1/x-1)z)} \\ = n! [z^n] \frac{1-x}{1-x\exp((1/x-1)zx)} \\ = n! [z^n] \frac{1-x}{1-x\exp((1-x)z)}.$$
The LHS is the same as the RHS and we have the claim.
Addendum. We have
$$n! [z^n] [w^k] \frac{w-1}{w-\exp((w-1)z)} \\ = n! [z^n] [w^{k+1}] \frac{w-1}{1-\exp((w-1)z)/w} \\ = n! [z^n] [w^{k+1}] (w-1) \sum_{q\ge 0} \frac{1}{w^q} \exp(q(w-1)z) \\ = [w^{k+1}] \sum_{q\ge 0} \frac{1}{w^q} q^n (w-1)^{n+1} = \sum_{q\ge 0} [w^{k+1+q}] q^n (w-1)^{n+1} \\ = (-1)^{n-k} \sum_{q=1}^{n-k} (-1)^q q^n {n+1\choose k+1+q}.$$
This justifies that $\left\langle n \atop k \right\rangle = 0$ when $k\ge n$ and hence the two coefficient extractors combined return zero in that case as claimed.