How to merge a $1$-cycle with an $n$-cycle

Question

I know that $(14625)$ is equal to $(14)(23)(36)(53)(54)$. I want to know how to get from one to the other algebraically - in other words how to 'absorb' $(3)$ into $(14625)$. The best I have managed is:

$$(14625)(3)$$ $$=(14)(4625)(3)$$ $$=(14)(6254)(3)$$ $$=(14)(625)(54)(3)$$ $$=(14)(256)(54)(3)$$ $$=(14)(56)(26)(54)(3)$$ $$=(14)(56)(26)(54)(36)(63)$$ $$=(14)(56)(26)(36)(63)(54)$$ $$=(14)(33256)(54)$$ $$=(14)(3256)(54)$$ $$=(14)(2563)(54)$$ $$=(14)(36)(53)(23)(54)$$

which is not correct. (The transpositions are correct but they are in the wrong order). I don't know what I did wrong.

Answer 1

From the comments:

my reasoning in that step was that $(ad)(bd)(cd) = (cbad)$, so $(56)(26)(36)(63) = (56)(26)(36)(36) = (33256)$.

This rule only works when $a, b, c, d$ are all different, as you can see.

Can you suggest an alternative method to get from $(14625)(3)$ to $(14)(23)(36)(53)(54)$?

One failsafe way is to write $$ (14625) = (14)(14)(14625) = (14)(23)(23)(14)(14625) = ... = (14)(23)(36)(53)(54)(54)(53)(36)(23)(14)(14625), $$ and then just multiply all the terms on the right until they all cancel.