Show that $C^*(\mathbb Z)$ is the universal $C^*$-algebra generated by a unitary. I.e.,show that for any unital $C^*$-algebra $A$ containing a unitary $u$, there is a unique unital *-homomorphism $\rho :C^*(\mathbb Z)\to A$ with $\rho (\delta_1) = u$.
The uniqueness part is done but unknown about existence part.
Existence: For any unital $C^∗$-algebra $$ containing a unitary $$, there exists a unital *-homomorphism $\rho \colon ^∗(\mathbb Z) \to $ such that $\rho(\delta_1) = $.
Uniqueness: The homomorphism $\rho$ is unique, meaning that there is no other homomorphism that satisfies the same condition.
Well, almost by definition $C^*(G)$ has the following universal property: given a unitary representation $(H,u)$ of $G$ i.e. a Hilbert space $H$ and a group homomorphism $u:G\to U(H)$, $g\mapsto u_g$, where $U(H)$ is the group of unitary operators over $H$, then there exists a unique $*$-homomorphism $\pi^u:C^*(G)\to B(H)$ s.t. $\pi^u(\delta_g)=u_g$. Note moreover that, since $C^*(G)=\overline{\text{span}}\{\delta_g:g\in G\}$, the image of $\pi^u$ is actually contained in the $C^*$-algebra generated by $\{u_g:g\in G\}$.
Now suppose that $A$ is a unital $C^*$-algebra and let $u\in A$ be a fixed unitary operator. Embed $A$ unitally inside $B(H)$ and consider the unitary represtantion $\mathbb{Z}\to U(H)$, with $n\mapsto u^n$. By the universal property of the full group $C^*$-algebra, there exists a unique $*$-homomorphism $C^*(\mathbb{Z})\to B(H)$ s.t. $\delta_1\mapsto u$. Note actually that the range of this $*$-homomorphism is contained in $C^*(u)$, and thus contained in $A$.