Hmm, I cannot see the problem which came up in the comments. If we assume simply an error in the notation and that the actually the Stirling numbers 2'nd kind are meant (as verbally exposed in the question) then the identity holds. This can even be checked simply using negative integer $x$ and computing Eulersums of appropriate order.
Let's write S2 the matrix of that numbers as
$ \qquad \small
\begin{array} {rrrrrrr}
1 & . & . & . & . & . & . & . \\
1 & 1 & . & . & . & . & . & . \\
1 & 3 & 1 & . & . & . & . & . \\
1 & 7 & 6 & 1 & . & . & . & . \\
1 & 15 & 25 & 10 & 1 & . & . & . \\
1 & 31 & 90 & 65 & 15 & 1 & . & . \\
1 & 63 & 301 & 350 & 140 & 21 & 1 & . \\
1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1 \\
\ldots &
\end{array}
$
where we use zero-based row- and columnindexes.
Then the problem can be restated as summing by building the dot product of one column by one row vector $V(x) = [1,x,x^2,x^3,...]$ with manageable (ideally infinite) dimension.
The numbers along a column can be seen as composed by finite compositions of geometric series.
Column 0 is $[1,1,1,1,1,...]$ and the dot-product with the V(x)-vector is then
$V(x)*S2[,0] = {1 \over 1-x}$
Column 1 is $[1-1,2-1,4-1,8-1,16-1,...]$ and the dot-product with the V(x)-vector is then
$V(x)*S2[,1] = {1 \over 1-2x} - {1 \over 1-x} = { (1-1x) - (1-2x) \over (1-1x)(1-2x) } = {x \over (1-1x)(1-2x) }$
One needs the simple composition of the other columns (see for instance in wikipedia) to see more examples for that decompositions, and also a general description for that compositions (where the text is:"Another explicit expanding of the recurrence-relation(...)").
I think the idea behind that homework-assignment was, that the student should find the compositions of powers such that the problem is ocnverted to describe the (finite) composition of closed forms of geometric series.
explain this "identical boxes"
When we put distinguishable objects into "identical boxes", it means we put distinguishable objects into distinguishable boxes, but then we want to consider rearranging the boxes to arrive at an equivalent arrangement, and we count the number of equivalence classes.
A better way of thinking about it is that we have a set of boxes (not a list), and each box is a nonempty set. So the order of the boxes doesn't matter, and the order of objects within each box doesn't matter.
Perhaps an example will help.
The number of ways to put $n$ distinguishable objects $1,2,3,\ldots, n$ into $k$ identical boxes when $n = 4$ and $k = 2$ is $7$:
Box 1 Box 2
123 4
124 3
134 2
234 1
124 3
12 34
13 24
14 23
1 how to construct the formula for it (the recursive one) preferably using bijection principle and just using "box and objects language"
Let's prove the recurrence relation
$$
\newcommand{\stir}{\genfrac\{\}{0pt}{}}
\stir{n+1}{k} = k \stir{n}{k} + \stir{n}{k-1}.
$$
Consider an arrangement of $n+1$ objects $(1,2,3,\ldots,n+1)$ put in $k$ boxes, of the kind counted by $\stir{n+1}{k}$.
There are two cases to consider: either object $n+1$ stands in its own box, or it shares its box with some other objects.
If it stands in its own box, then we have to distribute the remaining $n$ objects into the remaining $k-1$ boxes, and there are $\stir{n}{k-1}$ ways to do so. If not, then we have to put $n$ remaining objects into $k$ boxes, one of them distinguished from the rest, such that each box is nonempty (the box that contained object $n+1$ has to be nonempty too, as that is the case we are considering). The number of ways to do this is $k \cdot \stir{n}{k}$: first $\stir{n}{k}$ ways to put the $n$ objects into $k$ identical boxes, and then $n$ ways to choose which box to select as the distinguished one, which we will put $n+1$ into. (Note that nonempty is key here: because all boxes are nonempty, despite being indistuishable boxes they are distinguishable once any object is put in them; thus, we cannot put object $n+1$ into two different boxes and end up with the same resulting arrangement.)
2 explain the same but now with " set partition language"
In a partition of $1,2,3,\ldots,n+1$ into $k$ sets, there are two possibilities: either $\{k\}$ is an element of the partition, or $\{k\}$ is not an element of the partition. In the forms case,
there is a one-to-one, onto map (bijection) from all such partitions of $1,2,3,\ldots,n+1$ into $k$ parts to all partitions of $1,2,3,\ldots,n$ into $k-1$ parts, given by
(for $P$ a partition of $\{1,2,3,\ldots,n+1\}$
$$
P \quad\mapsto\quad P \setminus \{\{n+1\}\}.
$$
In the latter case, there is a $k$-to-one, onto map from all such partitions of $1,2,3\ldots,n+1$ into $k$ parts to all partitions of $1,2,3, \ldots, n$ into $k$ parts, given by
$$
P \quad\mapsto\quad \{x \setminus \{n+1\} \;|\; x \in P\}.
$$
3 define and construct the explicit formula for it
Explicit formulas are not always that nice :)
You can find a combinatorial answer to building an explicit formula here.
Best Answer
A set partition into $k$ parts is unordered partition of a set $S$ in unordered collection pairwise disjoint sets $S_1,S_2,...,S_k$. Number of set partitions of ${1,2,...,n}$ into $k$ parts is $S(n,k)$. Also a set partition of ${1,2,...,n}$ is equivalent to $ \cdot \times Seq\{1\} \times \cdot \times seq\{1,2\} \times \cdot \times \ldots \times seq\{1,2,\ldots,k\}$ where $\cdot$ and all numbers $j$ have size $1$. To see this if a number $l$ is at position $j$ then you put $j$ in $S_l$ and if $i^{th}$ dot is at position $j$ then you put $j$ in $S_i$. So $$ \sum_{n\geq 0 } S{(n,r)}x^n = \dfrac{x}{1-x}\times\dfrac{x}{1-2x}\times\ldots\times\dfrac{x}{1-rx}$$ Extracting coefficient of $x^n$ you get your formula.