Assume $a \in A$ where $A$ is a unital $C^*$-algebra. If $\lVert a \rVert = 1$ and $a^*=a$ we know that $1 \in \sigma(a)$, the spectrum of $a$. This follows from the fact that $\lVert a \rVert = r(a) = \sup_{\lambda \in \sigma(a)} \lvert \lambda \rvert$.
Is there a proof, which does not rely on the spectral radius formula ?
Suppose that $a$ is positive, that $\|a\|=1$ and that $a-1$ is invertible. This means that $0\not\in\sigma(a-1)$, so $1\not\in\sigma(a)$. As the spectrum is closed, there exists $\varepsilon>0$ such that $(1-\varepsilon,1+\varepsilon)\cap\sigma(a)=\varnothing$.
Consider the function $$f(t)=\begin{cases}t,&\ 0\leq t\leq 1-\varepsilon,\\ 1-\varepsilon,&\ 1-\varepsilon\leq t\end{cases} $$ On $\sigma(a)$, we have $f(t)=t$, so $f(a)=a$. Then, since $f(t)\leq1-\varepsilon$, $$ a=f(a)\leq (1-\varepsilon)\,I $$ It follows that $\|a\|\leq\|(1-\varepsilon)\,I\|=1-\varepsilon,$ a contradiction.
(the fact that for positive $a,b$, if $a\leq b$ then $\|a\|\leq\|b\|$ can be obtained in several ways; for instance by using that the norm of an element of a C$^*$-algebra can be obtained using the states; another easy way is representing on $B(H)$)