That formulation seems fine. What you've written is that $\sigma$ and $\tau$ are disjoint if nontrivial orbits of $\tau$ and $\sigma$ have no intersection.
Another way to think about this: let $X=\{1,\ldots,n\}$. We can express $\sigma=(i_1\,\cdots\,i_s)$ and $\tau=(j_1\,\cdots\, j_t)$, where $i_k,j_l\in X$ are numbers. Then $\sigma$ and $\tau$ are disjoint if and only if $i_k\neq j_l$ for all $k,l$.
The number of cycles in a permutation is equal to the number of orbits it has (meaning, the number of orbits of the cyclic subgroup it generates). Without loss of generality $\sigma=(1\cdots m)$ and we can consider it acting on $\Bbb Z/m\Bbb Z$ via $x\mapsto x+1$ (every other orbit of $\sigma$ is a fixed point) so $\sigma^k$ acts by $x\mapsto x+k$.
The orbits of $x\mapsto x+k$ look like $\{x,x+k,x+2k,\cdots\}$ which are precisely the cosets $x+\langle k\rangle$ of the additive subgroup $\langle k\rangle=\{0,k,2k,\cdots\}$ of $\Bbb Z/m\Bbb Z$. Thus, the number of orbits is equal to $m/|k|$ where we let $|k|$ represent the additive order of $k$ mod $m$ (equivalently, the size of $\langle k\rangle$). At this point, for the purpose of the problem, it suffices to determine when $|k|=m\Leftrightarrow \langle k\rangle=\Bbb Z/m\Bbb Z\Leftrightarrow [k]\in(\Bbb Z/m\Bbb Z)^\times$ is a unit mod $m$, but we could also more generally prove the formula for the order $|k|$.
Units of $\Bbb Z/m\Bbb Z$ are precisely residues of integers coprime to $m$. Clearly any noncoprime integer will have a residue which is a zero divisor, and conversely multiplication by the residue of a coprime integer is an injective map on a finite set hence bijective, so $1$ has a preimage which must be the multiplicative inverse mod $m$.
Now for the more general formula, which isn't strictly part of the original problem. Note the additive subgroups of the ring $\Bbb Z/m\Bbb Z$ are ideals, which correspond to ideals in $\Bbb Z$ containing the ideal $m\Bbb Z$ by the fourth isomorphism theorem; moreover, since $\Bbb Z$ is a PID, so must its quotients be, so $\langle k\rangle=\langle d\rangle$ for some divisor $d\mid m$. In fact since we easily compute $\left(\frac{k}{(k,m)},m\right)=\frac{(k,m)}{(k,m)}=1$ (so $\frac{k}{(k,m)}$ represents a unit mod $m$), we know $k$ is associate to $k/\frac{k}{(k,m)}=(k,m)\mid m$ hence $\langle k\rangle=\langle (k,m)\rangle$ which has order $\frac{m}{(k,m)}$.
Hence this is the number of orbits of $\sigma^k$, and the number of cycles in $\sigma^k$ is $m/\frac{m}{(k,m)}=(k,m)$.
In an abstract algebra course, there is usually an exercise to prove $|\sigma^k|=|\sigma|/(k,|\sigma|)$ in the section on cyclic groups. I consider this to be kind of tricky to prove at this level using only arithmetic, it's like programming in a machine language instead of something higher. One time a professor and I spent ten minutes lamely trying to remember how to prove the formula with just arithmetic. But later in a course in abstract algebra one becomes acquainted with the rings $\Bbb Z/m\Bbb Z$, and when one has a toolkit involving not just gcds but units and associates and ideals as well, the problem (at least in my opinion) becomes not only tolerable but intuitive and fun.
Best Answer
Look at the contrapositive. If $\sigma^k$ is not an $r$-cycle, then it is the product of shorter cycles, and there are an $i\in\operatorname{dom}\sigma$ and an integer $\ell$ such that $2\le\ell<r$ and $(\sigma^k)^\ell(i)=i$. But then the identity permutation and $(\sigma^k)^\ell$ are distinct members of $\langle\sigma^k\rangle$ sending $i$ to $i$, while $\langle\sigma\rangle$ clearly does not have a non-identity permutation that fixes $i$. Thus, $\langle\sigma^k\rangle\ne\langle\sigma\rangle$.