Easy Proof Adjoint(Compact)=Compact I am looking for an easy proof that the adjoint of a compact operator on a Hilbert space is again compact. This makes the big characterization theorem for compact operators (i.e. compact iff image of unit ball is relatively compact iff image of unit ball is compact iff norm limit of finite rank operators) much easier to prove, provided that you have already developed spectral theory for C*-algebras. 
By the way, I'm using the definition that an operator $T\colon H \to H$ is compact if and only if given any [bounded] sequence of vectors $(x_n)$, the image sequence $(Tx_n)$ has a convergent subsequence. 
edited for bounded

Here is an alternative proof, provided that you know that an operator is compact iff it is the operator-limit of a sequence of finite-rank operators.
Let $T: H \to H$ be a compact operator. Then $T= \lim_n T_n$ where the limit is w.r.t. the operatornorm and $T_n$ is a finite rank operator. Using that the $*$-involution is continuous, we get
$$T^*= \lim_n T_n^*$$
where $T_n^*$ is also a finite rank operator for all $n$. Thus $T^*$ is the limit of finite-rank operators and it follows that $T^*$ is compact as well.
