Let $q_k(n)$ denote the number of distinct partitions of some $n \in \mathbb{N}$ with exactly $k$ parts. How can I show that: $$k! \cdot q_k(n) \leq \binom{n-1}{k-1}$$
There is a related inequality that I reasoned out which was $k! \cdot p_k(n) \geq \binom{n-1}{k-1}$, where $p_k(n)$ denotes the number of partitions of $n$ with exactly $k$ parts. The way I justified this one was by recognizing the right-hand side counts the number of ordered $k$-tuples of positive integer solutions $(x_1, x_2, \ldots, x_k)$ satisfying $$\sum_{i=1}^k x_i = n$$
By definition, $p_k(n)$ is the number of partitions of $n$ with $k$ parts, and order is not considered within a partition. By multiplying this by $k!$ we count each permutation of the $k$-sized partitions.
For some ordered $k$-tuple of positive integer solutions, there are at most $k!$ distinct $k$-tuples containing the same numbers. This only occurs when all $k$ numbers are distinct. The inequality $k! \cdot p_k(n) \geq \binom{n-1}{k-1}$ follows.
Is there some similar reasoning to reach the first inequality?
You can use essentially the same argument as in your proof for the related inequality. Each partition with distinct parts corresponds to exactly $k!$ distinct compositions of the number $n$ with $k$ parts. So there are at least $k! q_k(n)$ compositions of $n$ with $k$ parts, and as in your proof for the related inequality, we know that the number of compositions of $n$ with $k$ parts is $$ \binom{n - 1}{k - 1}. $$
i.e. If $x_1 < x_2 < \dots < x_k$ is such that $x_1 + x_2 + \dots + x_k = n$, so that $(x_1, x_2, \dots, x_k)$ is one of the partitions counted by $q_k(n)$, then this corresponds to $k!$ solutions to $$ y_1 + y_2 + \dots + y_k = n $$ because we can take $y_i = x_{\sigma(i)}$ where $\sigma$ is some permutation of $\{1, 2, \dots, k\}$, and there are $k!$ such permutations. This implies that there are at least $k! q_k(n)$ solutions to $$ y_1 + y_2 + \dots + y_k = n $$ but we know that there are exactly $$ \binom{n - 1}{k - 1} $$ solutions.