Let $G$ be a Lie group of dimension $n$, and let $\mu$ be a left-invariant $n$-form. I've already shown that the pullback under a right translation, $R_h^\ast \mu$ is also a left-invariant form and that there exists a group homomorphism $\lambda:G \to \mathbb{R}^\ast$ such that
$$R_h^\ast \mu = \lambda (h) \mu.$$
Now I need to show that $\lambda$ is smooth, but I'm confused as to how. For starters, we have that
\begin{align*} (R_h^\ast \mu)_g (X_1,\dots,X_n)=\mu_{gh}((dR_h)_g X_1,\dots, (dR_h)_g X_n)) \end{align*}
and thus
$$\lambda(h) \mu_g(X_1,\dots,X_n)=\mu_{gh}((dR_h)_g X_1,\dots, (dR_h)_g X_n)).$$
I'm not sure what to do now. What I want to say is, since the right-hand side is smooth as $h$ varies, then $\lambda(h)$ is also smooth. But I don't know what it means to say that $h \mapsto dR_h$ is a "smooth map", so I can't be sure of my reasoning.
The multiplication map $m\colon G\times G\to G$, $m(g,h) = gh$, is smooth. This means that it induces a smooth bundle map $Dm\colon T(G\times G)\to TG$. In particular, $R_h(g)=m(g,h)$ and so for $X\in\mathfrak g = T_eG$, we have $(dR_h)_e(X)=Dm_{(e,h)}(X,0)\in T_hG$ and $X(h)=(dR_h)_e(X)$ is a smooth vector field on $G$.
Let's fix a basis $X_1,\dots,X_n$ for $\mathfrak g$ with $\mu(e)(X_1,\dots,X_n)=1$. Since $(R_h^*\mu)(e) = \lambda(h)\mu(e)$, $$\lambda(h) = \mu(h)\big(X_1(h),\dots,X_n(h)\big)$$ is smooth.