I know that the following is true for all formula $\alpha_1, \alpha_2, \cdots, \alpha_n$,
$$\vdash (\alpha_1\to(\alpha_2\to(\cdots(\alpha_{n-1}\to\alpha_{n})\cdots))) \leftrightarrow ((\alpha_1\wedge \alpha_2\wedge\cdots\wedge\alpha_{n-1})\to \alpha_n) $$
So we can simplify $\alpha_1\to(\alpha_2\to(\cdots(\alpha_{n-1}\to\alpha_{n})\cdots))$ to $(\alpha_1\wedge \alpha_2\wedge\cdots\wedge\alpha_{n-1})\to \alpha_n$. Then, is there a similar result for $(\cdots ((\alpha_1 \to \alpha_2 )\to \alpha_3 )\cdots \to \alpha_n)$?
There isn't anything as nice as the
$$(\alpha_1\to(\alpha_2\to(\cdots(\alpha_{n-1}\to\alpha_{n})\cdots))) \Leftrightarrow ((\alpha_1\wedge \alpha_2\wedge\cdots\wedge\alpha_{n-1})\to \alpha_n)$$
equivalence, but what you get is:
$$((((\cdots \to \alpha_{n-3} )\ \to \alpha_{n-2} )\to \alpha_{n-1} )\ \to \alpha_n) \Leftrightarrow ((((\cdots \land \neg \alpha_{n-3}) \lor \alpha_{n-2} )\land \neg \alpha_{n-1} )\ \lor \alpha_n) $$
See how that works? The terms end up alternating between being negated or not, and the operators in between end up alternating between $\land$'s and $\lor$'s. And in the end, you still have the same nesting of parentheses, so in that respect nothing is 'gained' by this conversion. Oh, and the nature of the most 'inside' expression that relates $\alpha_1$ and $\alpha_2$ depends on whether $n$ is even or odd:
If $n$ is even, you end up with:
$$((((\cdots (\neg \alpha_1 \lor \alpha_2) \cdots \land \neg \alpha_{n-3}) \lor \alpha_{n-2} )\land \neg \alpha_{n-1} )\ \lor \alpha_n) $$
But if $n$ is odd, you end up with:
$$((((\cdots (\alpha_1 \land \neg \alpha_2) \cdots \land \neg \alpha_{n-3}) \lor \alpha_{n-2} )\land \neg \alpha_{n-1} )\ \lor \alpha_n) $$
Finally, here is an inductive proof for all this:
Base:
$n=1$: Then $((((\cdots \to \alpha_{n-3} )\ \to \alpha_{n-2} )\to \alpha_{n-1} )\ \to \alpha_n)$ corresponds to just $\alpha_1$, while $((((\cdots (\alpha_1 \land \neg \alpha_2) \cdots \land \neg \alpha_{n-3}) \lor \alpha_{n-2} )\land \neg \alpha_{n-1} )\ \lor \alpha_n) $ corresponds to just $\alpha_1$ as well. Check!
$n=2$: Then $((((\cdots \to \alpha_{n-3} )\ \to \alpha_{n-2} )\to \alpha_{n-1} )\ \to \alpha_n)$ corresponds to $\alpha_1 \to \alpha_2$, while $((((\cdots (\alpha_1 \land \neg \alpha_2) \cdots \land \neg \alpha_{n-3}) \lor \alpha_{n-2} )\land \neg \alpha_{n-1} )\ \lor \alpha_n) $ corresponds to $\neg \alpha_1 \lor \alpha_2$, and they are indeed equivalent. Check!
Step:
The inductive hypothesis is that the claim holds for some arbitrary number $k$, i.e. that:
$$((((\cdots \to \alpha_{k-3} )\ \to \alpha_{k-2} )\to \alpha_{k-1} )\ \to \alpha_k) \Leftrightarrow ((((\cdots \land \neg \alpha_{k-3}) \lor \alpha_{k-2} )\land \neg \alpha_{k-1} )\ \lor \alpha_k) $$
Now let's consider $k+1$:
$$(((((\cdots \to \alpha_{k-3} )\ \to \alpha_{k-2} )\to \alpha_{k-1} )\ \to \alpha_k) \to \alpha_{k+1})\overset{Inductive Hypothesis}\Leftrightarrow$$
$$((((\cdots \land \neg \alpha_{k-3}) \lor \alpha_{k-2} )\land \neg \alpha_{k-1} )\ \lor \alpha_k) \to \alpha_{k+1})\overset{Implication}\Leftrightarrow$$
$$(\neg (((\cdots \land \neg \alpha_{k-3}) \lor \alpha_{k-2} )\land \neg \alpha_{k-1} )\ \lor \alpha_k) \lor \alpha_{k+1})\overset{DeMorgan}\Leftrightarrow$$
$$((((\cdots \lor \alpha_{k-3}) \land \neg \alpha_{k-2} )\lor \alpha_{k-1} )\ \land \neg \alpha_k) \lor \alpha_{k+1})$$
Check!