See Herbert Enderton, A Mathematical Introduction to Logic (2nd ed Harcourt - 2001), page 50.
With only the $\land$ and $\rightarrow$ connectives, if the sentence symbols in our formula are assigned the value $\top$, then the entire formula is assigned the value $\top$.
We have to proof this by induction on the lenght of the formula; i.e. we have to show that for any wff $\alpha$ built up using only these connectives we have that :
in each valuation $v$ such that $v(p_i) = \top$, for each $p_i$ in $\alpha$, then $v(\alpha) = \top$.
The proof is trivial :
Basis
$\alpha$ is $p_1$; then, $v(p_1) = \top = v(\alpha)$.
Induction step
$\alpha$ is $\alpha_1 \land \alpha_2$ or $\alpha_1 \rightarrow \alpha_2$, where we assume by induction hypotheses, that :
if $v(p_i)=\top$ for each $p_i$ in $\alpha_1$ and $\alpha_2$, then $v(\alpha_1)=v(\alpha_2)=\top$.
It's enough to use truth-tables.
Having shown this, we have shown that with only the two connectives $\land$ and $\rightarrow$ we are not able to "produce" a formula that, when all its sentence letters evaluates to $\top$ (i.e.TRUE), it gives as result the value $\bot$ (i.e.FALSE).
But with the valuation $v_0$ such that :
$v_0(p_1)=v_0(p_2)=v_0(p_3)= \top$
the formula $\alpha := \lnot [(p_1 \rightarrow p_2) \rightarrow (p_2 \rightarrow p_3)]$
will have the value $\bot$.
Another way to prove it is based on :
the equivalence between : $p \rightarrow q$ and $\lnot (p \land \lnot q)$,
in classical logic : because we need Double Negation.
Using this equivalence, we may rewrite our formula as :
$(p_1 \rightarrow p_2) \land \lnot (p_2 \rightarrow p_3)$
and again as :
$\lnot (p_1 \land \lnot p_2) \land (p_2 \land \lnot p_3)$.
Now we may apply the above argument in terms of valuations; with $v_0(p_1)=v_0(p_2)=v_0(p_3)= \top$, we have that :
$[\lnot (\top \land \lnot \top) \land (\top \land \lnot \top)] \equiv [\lnot (\top \land \bot) \land (\top \land \bot)] \equiv (\lnot \bot \land \bot) \equiv (\top \land \bot) \equiv \bot$.
But we have the above result that with only the $\land$ and $\rightarrow$ connectives, if the sentence symbols in a formula are assigned the value $\top$, then the entire formula is assigned the value $\top$.
Thus, is not possible to find a formula with only $\land$ and $\rightarrow$ that is equivalent to the original one.