Assume we are in $S_7$. Let $\alpha^3=(1,2,3,4)$. How to solve for $\alpha$?
This is what I did:
$\alpha^3=(1234)$ implies that after $3$ transformations $1 \mapsto 2$. So, begin by letting $\alpha = (1,a,b,2)$. We also know that $2 \mapsto 3$. So, $\alpha = (1,a,3,2)$. And finally, since $3 \mapsto 4$, $\alpha = (1,4,3,2)$. It can easily be verified that that is a solution.
Edit:
This process didn't seem to work for the case when Actually, $\alpha^4 = (1,2,3,4,5,6,7)$ does work using the technique (see Arthur's answer). However, it still doesn't produce all solutions.
So, is there a general technique to solve the equation? Moreover, is the a general technique for solving $\alpha^n = (a_1,a_2,...,a_m)$ when $\alpha \in S_j$?
Where is the problem with using your approach? Starting with $(1, *,*,*,*,*,*)$, since $1$ after four applications becomes $2$, we must have $(1,*,*,*,2,*,*)$. Further applications give, in order $$ (1,3,*,*,2,*,*)\\ (1,3,*,*,2,4,*) $$ And so on. At last, we get $(1, 3, 5, 7, 2, 4, 6)$.
In general, it's not so easy to calculate, and there are often several solutions. Especially if the disjoint cycles in $\alpha^n$ have length that divide $n$, such as $$\alpha^4 = (12)(34)(56)(78)$$ in $S_8$.