Is the following argument?
Suppose $V_1,\dots,V_m$ are vector spaces. Prove that $\mathcal{L}(V\times\cdots\times V_m,W)$ and $\mathcal{L}(V_1,W)\times\cdots\times\mathcal{L}(V_m,W)$ are isomorphic vector spaces.
Proof. We define $\Delta:\mathcal{L}(V\times\cdots\times V_m,W)\to\mathcal{L}(V_1,W)\times\cdots\times\mathcal{L}(V_m,W)$ such that $\Delta T = (T_1,T_2,\dots,T_m)$ where $$\begin{cases} T_1v = T(v,0,0,\dots,0),\forall v\in V_1\\ T_2v = T(0,v,0,\dots,0),\forall v\in V_2\\ \vdots\\ T_mv = T(0,0,0,\dots,v),\forall v\in V_m\\ \end{cases}$$ Now let $S,T\in\mathcal{L}(V\times\cdots\times V_m,W)$ and assume $\Delta S = (S_1,S_2,\dots,S_m)= (T_1,T_2,\dots,T_m)= \Delta T$ consequently we have $T_k=S_k,\forall k$ thus given an arbitrary $(v_1,v_2,\dots,v_m)\in V_1\times\cdots\times V_m$ we have $T_kv_k = S_kv_k,\forall k$, then by considering the above definition, the following becomes immediately apparent. $$\begin{cases} T(v_1,0,0,\dots,0) = S(v_1,0,0,\dots,0)\\ T(0,v_2,0,\dots,0) = S(0,v_2,0,\dots,0)\\ \vdots\\ T(0,0,0,\dots,v_m) = S(0,0,0,\dots,v_m) \end{cases}$$ This when taken together with the linearity of $S$ and $T$ then implies that $S(v_1,v_2,\dots,v_m) = T(v_1,v_2,\dots,v_m)$, consequently $S=T$, thus $\Delta$ is injective.
Now let $(T_1,T_2,\dots,T_m)\in\mathcal{L}(V_1,W)\times\cdots\times\mathcal{L}(V_m,W)$. and define $T\in\mathcal{L}(V\times\cdots\times V_m,W)$ such that $T(v_1,v_2,\dots,v_m) = T_1v_1+T_2v_2+\cdots+T_mv_m$, then given the above definition of $\Delta$ it is apparent that $\Delta(T) = (T_1,T_2,\dots,T_m)$.
We will leave it to the reader to show that $\Delta$ is infact linear, this together with above argument demonstrates that $\Delta$ is an isomorphism and $\mathcal{L}(V\times\cdots\times V_m,W)$ and that $\mathcal{L}(V_1,W)\times\cdots\times\mathcal{L}(V_m,W)$ are isomorphic.
$\blacksquare$