Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$.
If $\sigma(M) = 2M$, then $M$ is said to be perfect.
Currently, as of December 2018, there are $51$ known examples of even perfect numbers -- on the other hand, we still do not know whether there are any odd perfect numbers.
Euler derived the general form that an odd perfect number $N$ must take:
$$N = {q^k}{n^2},$$
where $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$. We call $q$ the special or Euler prime of $N$.
Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$.
In [Dris, 2012], it was shown that the implications
$$n < q \Longrightarrow k = 1$$ and $$n < q^2 \Longrightarrow k = 1$$
are true.
Now, note that, since $q$ and $\sigma(q) = q + 1$ are consecutive integers, then the following implications are true.
Case 1: $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2: $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
The remaining cases to be considered are:
Case 3: $q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$
Case 4: $n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$
Case 5: $n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$
Case 6: $n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$
Am I right in thinking that it is possible to prove the following implication, given the above cases?
CLAIM: If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $$n < q^{k + 1} \Longrightarrow k \neq 1.$$
Easy to prove for Cases 1 and 2. Should also be doable for the rest. Anybody have any ideas?
Added September 12 2016
From one of the answers below, it can be shown that the condition $q^2 < n$ is equivalent to the truth of the implication $$n < q^{k + 1} \Longrightarrow k \neq 1.$$