Here is one way of looking at the answer to this problem.

For any function $f: \mathbb{N} \to \mathbb{R}$, one can form the exponential generating function or formal power series $E_f(x) = \sum_{j \geq 0} \frac{f(j) x^j}{j!}$. The power series product of such egf's corresponds to the convolution product defined by

$$(f \ast g)(n) = \sum_{j+k = n} \frac{n!}{j!k!} f(j) g(k).$$

For example, if $g(k) = (-1)^k$, then $E_g(x) = \exp(-x)$, and the convolution product

$$h(n) := (f \ast g)(n) = \sum_j (-1)^{n-j} {n \choose j} f(j)$$

corresponds to the equation $E_h = E_f \cdot \exp(-x)$. Rewriting this as $E_h \exp(x) = E_f$ (and with an option to swap out other dummy variables for $j$), we arrive at a Möbius inversion formula

$$h(n) = \sum_k (-1)^{n-k} {n \choose k} f(k) \qquad \Leftrightarrow \qquad f(k) = \sum_j {k \choose j} h(j).$$

In the above problem, we are thus (up to a sign $(-1)^n$) trying to solve for $h(j)$ in the formula

$$k^{m+n} = \sum_n {k \choose j} h(j),$$

but notice that we can interpret $k^{m+n}$ combinatorially as the number of functions from an $(m+n)$-element set to a $k$-element set. Each such $f$ can be uniquely factored as a surjection followed by a subset inclusion; here ${k \choose j}$ counts the number of $j$-element subsets, and correspondingly $h(j)$ is counting the number of surjections from an $(m+n)$-element set to a $j$-element set $[j] := \{1, 2, \ldots, j\}$. Such a surjection can be regarded as a partition of $[m+n]$ into $j$ nonempty classes together with a labeling of the classes by the elements $1, \ldots, j$, and there are $j!$ such labelings. The second Stirling number $S(m+n, j)$ is in fact combinatorially defined as the number of partitions on $[m+n]$ with $j$ classes, and so we have shown

$$h(j) = S(m+n, j) \cdot j!$$

or $h(n) = S(m+n, n) \cdot n!$, as indicated in the answer above.

Here is one way of looking at the answer to this problem.

For any function $f: \mathbb{N} \to \mathbb{R}$, one can form the exponential generating function or formal power series $E_f(x) = \sum_{j \geq 0} \frac{f(j) x^j}{j!}$. The power series product of such egf's corresponds to the convolution product defined by

$$(f \ast g)(n) = \sum_{j+k = n} \frac{n!}{j!k!} f(j) g(k).$$

For example, if $g(k) = (-1)^k$, then $E_g(x) = \exp(-x)$, and the convolution product

$$h(n) := (f \ast g)(n) = \sum_j (-1)^{n-j} {n \choose j} f(j)$$

corresponds to the equation $E_h = E_f \cdot \exp(-x)$. Rewriting this as $E_h \exp(x) = E_f$ (and with an option to swap out other dummy variables for $j$), we arrive at a Möbius inversion formula

$$h(n) = \sum_k (-1)^{n-k} {n \choose k} f(k) \qquad \Leftrightarrow \qquad f(k) = \sum_j {k \choose j} h(j).$$

In the above problem, we are thus, modulo a sign $(-1)^n$, trying to solve for the (unique) $h(j)$ satisfying

$$k^{m+n} = \sum_j {k \choose j} h(j),$$

but notice that we can interpret $k^{m+n}$ combinatorially as the number of functions from an $(m+n)$-element set to a $k$-element set. Each such $f$ can be uniquely factored as a surjection followed by a subset inclusion; here ${k \choose j}$ counts the number of $j$-element subsets, and correspondingly $h(j)$ must be counting the number of surjections from an $(m+n)$-element set to a $j$-element set $[j] := \{1, 2, \ldots, j\}$. Such a surjection can be regarded as a partition of $[m+n]$ into $j$ nonempty classes together with a labeling of the classes by the elements $1, \ldots, j$, and there are $j!$ such labelings. The second Stirling number $S(m+n, j)$ is in fact combinatorially defined as the number of partitions on $[m+n]$ with $j$ classes, and so we have shown

$$h(j) = S(m+n, j) \cdot j!$$

or $h(n) = S(m+n, n) \cdot n!$, as indicated in the answer above.