Why is $U(\Lambda)^{-1} = U(\Lambda^{-1})$ for a unitary representation? This is from the beginning of Srednicki's QFT textbook, where he writes (approximately): 
In QM we associate a unitary operator $U(\Lambda)$ to each proper orthochronous Lorentz transformation $\Lambda$. These operators must obey the composition rule 
$$U(\Lambda'\Lambda) = U(\Lambda')U(\Lambda).$$
So far OK. 
But where does he get the following from?
$$U(\Lambda)^{-1} = U(\Lambda^{-1})$$
 A: I'd like to add a something to V. Moretti's (correct) answer.  You may be wondering

Where does the property $U(I) = I$ come from?

This property and the original one you asked about, generally hold for any group representation and, in fact, for any group homomorphism.
Let $\mathrm{SO}(1,3)^+$ denote the proper, orthochronous Lorentz group, and let $U(\mathcal H)$ denote the group of unitary operators on the Hilbert space $\mathcal H$ of the theory.  The representation $U:\mathrm{SO}(1,3)^+\to U(\mathcal H)$ about which you are asking, is a group representation which, in particular, is a group homomorphism.  A group homomorphism from a group $G$ to a group $H$ is a mapping $\phi:G\to H$ such that
\begin{align}
  \phi(g_1g_2) = \phi(g_1)\phi(g_2)
\end{align}
for all $g_1, g_2\in G$.  This property implies some interesting things.  Let $I_G$ be the identity of $G$ and $I_H$ be the identity of $H$.  Then notice, for example, that
\begin{align}
  \phi(I_G) =\phi(I_GI_G) = \phi(I_G)\phi(I_G) ,
\end{align}
and multiplying by $\phi(I_G)^{-1}$ on both sides gives
\begin{align}
  \phi(I_G) = I_H.
\end{align}
In other words

Group homomorphisms map the identity to the identity.

V. Moretti's answer can then be adopted in general to show that they also have the property $\phi(g^{-1}) = \phi(g)^{-1}$.
A: Indeed, it is also necessary to assume that $U(I)= I$ where the former $I$ is the identity matrix in the group and the latter is the identity operator in the Hilbert space.
So, taking advantage of your first identity you have:  $$U(\Lambda^{-1})U(\Lambda) = 
U(\Lambda)U(\Lambda^{-1}) =  U(I)=I$$ and it implies $U(\Lambda^{-1})= U(\Lambda)^{-1}$ because of uniqueness of the inverse operator.
