Suppose $|K|=\kappa, |L|=\lambda, |M|=\mu$ and $L \cap M=\emptyset$. Prove that $$(\kappa^{\lambda})^{\mu}=\kappa^{\lambda \cdot \mu}$$ My attempt:
Suppose $F : K^{ L \times M} \rightarrow (K^L)^M$. Let $f: L \times M \rightarrow K$. Then we define $F(f)=g$ where $g:M \rightarrow K^L$ and for all $m \in M$, we have $g(m)=h$ where $h: L \rightarrow M$ and $h(l)=f(l,m)$ for all $l \in L$.
Aim: Prove $F$ is bijective.
Surjective: Let $g : M \rightarrow K^L$. Then $g(m)=h$ for all $m \in M$. Also $h(l)=k$ for all $l \in L$. Let $f : L \times M \rightarrow K$ such that $f(l,m)=k$. The problem here is: I don't know whether such $f$ exists or not.
Injective: Suppose $F(f)=F(f^{\prime})$. Then I don't know how to proceed from here.
If $F(f)=F(f')$ then $$[F(f)](m)=[F(f')](m)$$ for each $m$. Since $F(f)(m)$ is a function from $L$ to $K$, you get $$[[F(f)](m)](l)=[[F(f')](m)](l)$$ for each $m\in M$, $l\in L$. But by definition, $[[F(f)](m)](l)=f(m,l)$ so $f=f'$.