If $q^k n^2$ is an odd perfect number with Euler prime $q$, then it is known that $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right),$$ where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$ and $D(x)=2x-\sigma(x)$. (See this paper for a proof.)
Here is my question:
Does $\sigma(n^2)/q^k < q$ hold?
MY ATTEMPT
It is easy to prove that $\sigma(n^2)/q^k \neq q$. For suppose to the contrary that $$q=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$ This means that $$2n^2 = q\sigma(q^k)$$ which implies that $q \mid n^2$, contradicting $\gcd(q,n)=1$.
So now, assume that $\sigma(n^2)/q^k > q$. We obtain $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right)>q,$$ from which we get $$2n^2 > \sigma(n^2) > q^{k+1}$$ $$2n^2 > q\sigma(q^k)$$ $$2n^2 - \sigma(n^2) > q\sigma(q^{k-1}) \geq q$$ $$\gcd(n^2, \sigma(n^2)) > q,$$ whence I do not arrive at a contradiction.
If the conjecture $q^k < n$ is true (see here (page 117), here, and here), then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we obtain $$\sigma(n^2) < q\cdot{q^k} < n\cdot{n} = n^2$$ which is a contradiction.
Also, if the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we have $$\sigma(n^2) < q\cdot{q^k} = q\cdot{q} < n\cdot{n} = n^2$$ which again is a contradiction.
Hence, conjecturally we expect $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right) > q$$ to be true.