Let $\cal{H}$ be a separable Hilbert space and $\cal{K(\cal{H})}$ the algebra of compact operators acting on $\cal{H}$. Then $$\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})\cong\cal{K}(\cal{H}\otimes H).$$
How to prove it? I can find a mapping $\phi$ from $\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})$ to $\cal{B}(H\otimes\cal{H})$, but there is some difficult to prove the range of $\phi$ is justly $\cal{K}(\cal{H}\otimes H)$.
P.S.: Is $\mathbb{M}_n(\mathbb{C})\otimes\mathbb{M}_n(\mathbb{C})\cong\mathbb{M}_{n^2}(\mathbb{C})$?
Thanks a lot!