Here's a nice proof that uses group actions.
Let $x_1,\ldots,x_n$ be $n$ unknowns, and consider
$$\Delta= \prod_{1\leq i\lt j\leq n} (x_j-x_i).$$
For example, for $n=4$, we would have
$$\Delta = (x_2-x_1)(x_3-x_1)(x_4-x_1)(x_3-x_2)(x_4-x_2)(x_4-x_3).$$
Given a permutation $\sigma\in S_n$, define a function $f_{\sigma}\colon \{\Delta,-\Delta\} \to \{\Delta,-\Delta\}$ by letting
$$f_{\sigma}(\Delta) = \prod_{1\leq i\lt j\leq n} (x_{\sigma(j)}-x_{\sigma(i)}),$$
and $f_{\sigma}(-\Delta)=-f_{\sigma}\Delta$.
Note that since $\sigma$ is a permutation, $f_{\sigma}(\Delta)=\Delta$ or $f_{\sigma}(\Delta) = -\Delta$. Also, if $\sigma,\rho$ are two permutations, then $f_{\sigma}\circ f_{\rho} = f_{\sigma\rho}$, as is easy to verify.
Now, let's consider what a transposition $\tau=(a,b)$ does to $\Delta$. Without loss of generality, say $a\lt b$.
The factors $(x_j-x_i)$ with neither $i$ nor $j$ equal to $a$ nor $b$ are unchanged.
For the pairs with exactly one index in $\{a,b\}$, we have two classes: those in which the other index is between $a$ and $b$, and those where the other index is not between $a$ and $b$.
If the other index is between $a$ and $b$, then $x_j-x_a$ is sent to $-(x_b-x_j)$ and $x_b-x_j$ is sent to $-(x_j-x_a)$; the two sign changes cancel each other out.
If the other index is larger than $b$, then $x_j-x_a$ and $x_j-x_b$ are swapped, with no sign changes.
If the other index is smaller than $a$, then $x_a-x_i$ and $x_b-x_i$ are swapped, with no sign changes.
Finally, the factor $x_b-x_a$ is sent to $-(x_b-x_a)$.
In summary, if $\tau$ is a transposition, then $f_{\tau}(\Delta)=-\Delta$, $f_{\tau}(-\Delta)=\Delta$.
Now take an arbitrary permutation $\sigma$, and express it as a product of transpositions in two different ways:
$$\sigma = \tau_1\cdots \tau_r = \rho_1\cdots\rho_s.$$
Then
$$f_{\sigma}(\Delta) = f_{\tau_1\cdots\tau_r}(\Delta) = f_{\tau_1}\circ\cdots\circ f_{\tau_r}(\Delta) = (-1)^r\Delta$$
and
$$f_{\sigma}(\Delta) = f_{\rho_1\cdots\rho_s}(\Delta) = f_{\rho_1}\circ\cdots\circ f_{\rho_s}(\Delta) = (-1)^s\Delta.$$
Therefore, $(-1)^r\Delta = (-1)^s\Delta$, so $r$ and $s$ have the same parity: both odd, or both even.
Best Answer
First, notice that, since 3-cycles are even, and products of even permutations are even themselves (if you know what it means, this comes from the fact that the map sending each permutation to its sign is a homomorphism), for a permutation to be written as a product of 3-cycles you need it to be even. Luckily, your $\alpha$'s are.
Now, let me give you the general argument here, that hopefully you will be able to specialize to the cases above. Take an even permutation $\sigma$ and write it as a product of transpositions, say $\tau_1\cdots\tau_k$, where $k$ must be even. Looking at $\tau_1\tau_2$, you can suppose $\tau_1\neq\tau_2$, otherwise their product is the identical permutation. Then, you have two cases:
Iterating and using the fact that $k$ is even, you get a full decomposition of $\sigma$ in 3-cycles. Incidentally, this actually proves that any even permutation can be written as a product of 3-cycles (again, provided you are familiar with the terminology, $A_n$ is generated by 3-cycles).