You are looking at permutations of $n$ and each cycle contributes a factor $x$.
Now, when you place the first element, there is only one possibility, it starts a cycle, this gives $x$.
When you place the second element, it either starts a new cycle giving $x$ or it is placed in the cycle of the first element, this gives $x+1$
When you place element $k$, it either starts a new cycle giving $x$ or it is placed as image of one of the elements that are already there, this gives $x+k-1$.
So, in total, you get the product on the right-hand side.
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.
Best Answer
Suppose you have the sequence $2, 5, 8, 11,...$
One idea might be to have the terms in the sequence as the coefficients of some polynomial. This polynomial is called a generating function of the sequence.
For example:
$$ g(x)=2x^1+5x^2+8x^3+11x^4+...$$
All seems fine but how do you recover the sequence if you have the generating function? In this example it is obvious but let's try out a couple of ideas to gain a greater insight.
Notice that differentiation of a power reduces its power by one. We are going to use this idea as a way of sliding the terms of the sequence forwards.
Suppose we want the first term we differentiate $g(x)$ once.
$$g'(x)=2+10x+24x^2+44x^3+....$$
Notice now we can evaluate the polynomial at $x=0$ and find the first term! $g'(0)=2$
The second term is more troublesome, can you see why?
Note the second derivative of the function is $g''(x)=10+48x+132x^2+...$ and that $g''(0)=10$ which is not the 2nd term. All is not lost as we can divide $g''(0)$ by $2$ to recover the second term (we multiplied by $2$ in differentiating). This problem will be exaggerated if we want higher terms in the sequence. The 4th term is found by computing $g^{iv}(0)$ and dividing by $4 \times 3 \times 2$ or $4!$
This situation is a pain and so we may choose to adapt the method.
One idea might be to preload the generating function with the dividing factorials first.
Let's define $$G_1(x)=2\frac{x^1}{1!}+5\frac{x^2}{2!}+8\frac{x^3}{3!}+11\frac{x^4}{4!}+...$$
Now we may differentiate the appropriate number of times and substitute $x=0$ with out hassle.
Finally we can observe the connection to the exponential function as $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
Hope this helps to demystify the exponential at least in part.