See Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001).
The Unique Readibility Theorem [page 40] says that for any formula $\varphi$, its immediate predecessors (that are well-formed formulae !) are uniquely determined.
This means that each formula, written without "colloquialisms" : e.g. as $(\alpha \land \beta)$ and not $\alpha \land \beta$, is "built-up from the set of sentence symbols by the five operations in a unique way.
Consider the formula $\varphi$ and assume that we can decompose it in two different way :
$\varphi := (\alpha \land \beta)$ and $\varphi := (\gamma \land \delta)$, where either $\alpha$ and $\gamma$ are different strings, or $\beta$ and $\delta$ are different strings, or both.
Consider the cases :
(i) $\alpha$ and $\gamma$ are different.
Both $\alpha$ and $\gamma$ are proper initial segments of $\varphi$ (i.e. initial substrings of $\varphi$); thus, one must be a proper initial segment of the other.
Then, say, $\alpha$ is a proper initial segment of $\gamma$ (of course, $\alpha$ is nonempty : otherwise our $\varphi$ is $(\land \beta)$, that it is not well-formed).
By Lemma 13B [page 30], being a proper initial segment of the wff $\gamma$, $\alpha$ has more left brackets than right ones; but $\alpha$ is also a wff and thus, by Lemma 13A [page 30], it has the same number of left and right brackets. Impossible!
The other case, $\gamma$ being a proper initial segment of $\alpha$ instead, is equally impossible.
(ii) If $\alpha$ and $\gamma$ match, this forces $\beta$ and $\delta$ to be the same
string, since the strings $(\alpha \land \beta)$ and $(\gamma \land \delta)$ are the same, and it's okay again.
Having answered "no" in all cases, we are done.