Inequalities on determinants of rational matrices. Here are two claims in a proof I'm reading that I don't get. Can I have some more elaboration?

Let $A=\left(\dfrac{p_{i j}}{q_{i j}}\right)_{i j}, i,j=1,...,n$, and
$\operatorname{det}(A)=\dfrac{p}{q}$ for completely reduced fractions.
$\dfrac{p_{i j}}{q_{i j}}, \dfrac{p}{q}$ and $q_{i j}, q>0 .$ We have:

*

*$ q+1 \leq \prod_{i, j=1}^{n}\left(q_{i j}+1\right)$.

*By the definition of the determinant via the Laplace formula: $|\operatorname{det}(A)| \leq \prod_{i, j=1}^{n}\left(\left|p_{ij}\right|+1\right)$.


 A: We know that $\prod_{i,j=1}^n q_{ij}$ is a common multiple of all the denominators in the determinant formula. This implies that there is some $m \in \mathbb{N^+}$ such that:
$$ mq = \prod_{i,j=1}^n q_{ij} $$
Because each $q_{ij}, q, m > 0$, the following inequality follows:
$$ q + 1\leq \prod_{i,j=1}^n \big(q_{ij} + 1\big) $$
For the second part, we can use induction on $n$:
$$ \sum_{i=1}^n (-1)^{n+i} \frac{p_{ni}}{q_{ni}} |C_{ni}| \leq \sum_{i=1}^n \Big(|p_{ni}| \prod_{j,k=1, j \neq n, k \neq i}^{n} \big(|p_{jk}| + 1\big)\Big) \leq \prod_{i=1,j=1}^n \big(|p_{ij}| + 1\big) $$
where $C_{ij}$ is the minor matrix in the Laplace formula.
Note that in the last step, we use the fact that $|p_{ij}| + 1 \geq 1 $ in order to simplify the expression:
$$  \prod_{j,k=1, j \neq n, k \neq i}^{n-1} \big(|p_{jk}| + 1\big) \leq \prod_{j,k=1, j \neq n}^{n-1} \big(|p_{jk}| + 1\big) $$
and then:
$$ \sum_{i=1}^n \Big(|p_{ni}| \prod_{j,k=1, j \neq n}^{n} \big(|p_{jk}| + 1\big)\Big) = \Big(\sum_{i=0}^n |p_{ni}|\Big)\prod_{j,k=1, j \neq n}^{n} \big(|p_{jk}| + 1\big)$$
We can further simplify this using the fact:
$$ \sum_{i=1}^n |p_{ni}| \leq \prod_{i=1}^n \big(1 + |p_{ni}|\big) $$
There's probably an easier way but this stood out to me since Laplace expansion is inductive by nature.
Edit:
Probably easier approach to the second part is just realising what the expansion of $ \prod(1 + |p_{ij}|)$ is and that the terms in $|\det(A)|$ ignoring all $q_{ij}$ form some subset of the terms in that expansion
