If $X$ is a Kan complex, then it is an easy consequence of the existence of the Quillen model structure or of the basic theory of anodyne extensions, that $X^{K}$ is also Kan. However, I am interested in a "direct" proof of this fact, appealing to neither of the above (which is apparently how these sorts of things were proved before Gabriel-Zisman).

Here is the argument I would like to complete:

Let $X$ be a Kan complex, and let $K$ be any simplicial set. It suffices to show that $Hom(K\times \Delta^{n},X) \rightarrow Hom(K \times \Lambda_{k}^{n},X)$ is surjective for all horn inclusions. Since $K\simeq \text{colim}_{m} sk_{m}(K)$ is a filtered colimit, the claim follows if $$Hom(sk_{m}(K)\times \Delta^{n},X) \rightarrow Hom(sk_{m}(K) \times \Lambda_{k}^{n},X)$$ is surjective for all $m \ge 0$. Proceed by induction. The base case is easy, but the induction step is what's giving me trouble. By the induction hypothesis, we can choose a solution

$$ h: sk_{m-1}K \times \Delta^{n} \rightarrow X$$

to the lifting problem

$\require{AMScd}$ $\begin{CD} sk_{m-1}K \times \Lambda^{n}_{k} @>>> sk_{m}K \times \Lambda^{n}_{k} @>f>> X \\ @VVV @VVV \\ sk_{m-1}K \times \Delta^{n} @>>> sk_{m}K \times \Delta^{n} \\ \end{CD}$

However, as the square above is not a pushout, we can't immediately get a solution to the lifting problem of interest. Now my thought is that maybe there's some way to directly construct the desired lift $g:sk_{m}K \times \Delta^{n} \rightarrow X$ from $f$ and $g$, since all we need to do is specify the images of the nondegenerate simplices of $sk_{m}K \times \Delta^{n}$ which are in neither $sk_{m-1}K \times \Delta^{n}$ nor $sk_{m}K \times \Lambda_{k}^{n}$. However, this is where I run into a wall, as I don't know a simple way to describe these.

Ideally, I would like to see this induction step completed. Does anyone have a solution to this or suggestions for possible solutions? It would be just as well if someone knows a different but equally "direct" argument, not appealing to the machinery mentioned above.