I've trying to multiply out this disjoint cycle in order to get the cycle from the computation $f^5$ but when I get to $f^4$ I just get stuck in a loop of 1 cycles from both the left and right cycle and I have no idea what to do next. Any advice on what I'm doing wrong would be appreciated, thanks
$f = (16)(96287) = (162879)$
$f^1 = (162879)(162879) = (127)(689)$
$f^2 = (127)(689)(162879) = (18)(67)(29)$
$f^3 = (18)(67)(29)(162879) = (172)(698)$
$f^4 = (172)(698)(162879) = \;(197826)$
$f^5 = (197826)(162879) = ????$
Very close. First, fixing your exponents, you should have:
$$\begin{align} f\ \ &= (162879)\\ f^2 &= (162879)(162879) = (127)(689)\\ f^3 &= (127)(689)(162879) = (18)(67)(29)\\ f^4 &= (18)(67)(29)(162879) = (172)(698)\\ f^5 &= (172)(698)(162879) = (197826)\\ f^6 &= (197826)(162879) = \textrm{????} \end{align}$$
Now, for the last one, it sounds like you were concerned about getting a product of $1$-cycles, but that's exactly what should happen: $$f^6 = (1)(2)(6)(7)(8)(9),$$ which is the identity permutation on those six elements. (Side note: perhaps your permutation is actually on nine elements, but you've been omitting the $1$-cycles for $3$, $4$, and $5$? In that case, $f^6 = (1)(2)(3)(4)(5)(6)(7)(8)(9)$. Either way, we could also just write something like $f^6 = \iota$ for brevity, where $\iota$ is the identity permutation.)
Note that $f$ is a $6$-cycle, so it is to be expected that after six applications, it returns to the identity.