Let the set $\hom(V_1, ..., V_p ; U) $ be the set of all p-linear maps from the space $V_1 \times ... \times V_p \to U$, then we have $$\dim ( \hom(V_1, ..., V_p ; U) ) = \dim(V_1) \cdot .... \cdot \dim(V_p) \cdot \dim(U)$$
I know that the maps in the space $\hom(V_1, ..., V_p ; U)$ are determined by by the images of the basis of the spaces $V_1 \times ... \times V_p $, so combinatorially one can easily prove the above theorem.However, I was wondering, how can we prove it by induction if it is possible ?
You could prove that for any $k$, $V_1,\dots, V_k$ and $U$, $\dim (\hom (V_1,\dots,V_k;U))=\dim(V_1) \cdot \dim (V_2,\dots,V_{k};U)$. Then you prove by (bounded) induction on $k$ that $\dim (\hom (V_1,\dots,V_p;U))=\left(\prod\limits_{i=1}^k \dim (V_k)\right)\dim (\hom (V_{k+1},\dots, V_p;U))$. The case $k=0$ is trivial, $k \implies k+1$ is the previous property, and $k=p-1$ gives you your result (up to $\dim (\hom(A;B))=(\dim A)\cdot (\dim B)$).
You could also prove the more general fact that $\hom (V_1,\dots,V_p;U)\cong \hom (V_1; \hom (V_2,\dots,V_p;U))=V_1\to_L \hom (V_2,\dots,V_p;U)$ (where $A\to_L B$ means $\hom (A;B)$). Then, you prove by induction that $\hom (V_1,\dots,V_p;U)\cong V_1\to _L \dots \to_LV_p \to_L U$. You recover the previous result by using the fact that $\dim (A\to_L B) = (\dim A)\cdot (\dim B)$. This property is stronger and I would therefore prefer this version but the fact that $\to_L$ is not associative makes notations a bit more annoying.
Another way would be to use tensor products. $\hom(V_1,\dots ,V_p;U)\cong\hom (V_1\otimes V_2,V_3,\dots, V_p;U)$. Then again, by induction, $\hom(V_1,\dots ,V_p;U)\cong \hom (V_1 \otimes \dots \otimes V_p;U)$ and you can conclude using $\dim (\hom (A;B))=(\dim A)\cdot (\dim B)$ and $\dim (A\otimes B)=(\dim A)\cdot (\dim B)$. The tensor product tends to be introduced a bit later in courses, but if you know it, I'd say that this is the "best" proof.