Distributivity of group action

Question

I have here groups $G$, $H$, a wreath product $G^n\wr H$ and a set $X$ of order $n.$ $H$ is acting (right action) on $X$ by permutation. To my understanding elements of $H$ are now also permuting components of elements of $G^n$ with the same action that $H$ acts on $X$ with, and we denote this $g.h$. This $g.h$ is now an element of $G^n$. How would I go about showing that $$(g_1g_2).h=(g_1.h)(g_2.h)$$ for all $g_1,g_2\in G^n,h\in H?$ Intuitively this seems to make sense, but is this a general property? My current attempt at this proof is as follows: $$(g_1\ast g_2).h=\phi((g_1\ast g_2,h))=\phi((g_1,h))\ast\phi((g_2,h))=(g_1.h)\ast(g_2.h),$$ where $(\ast)$ is the group operation in $G^n$ and $\phi$ is the action in question. My problem is specifically whether the second equality is valid or not, and if it is valid, where does it come from?