This might be a long-winded way of proving something so obvious, but I want to have it checked if it holds up.

Claim: There does not exist a sequence $\{{a}_{n \in \mathbb{N}}\}$ of natural numbers such that $a_{n+1} < a_{n}$ for every $n$.

Proof: This amounts to showing that the set $M = \{ m \in \mathbb{N} : m \leq a_{0}\}$, where $a_{0}$ is the first member from any sequence in infinite descent, is finite. Suppose for the sake of contradiction that there exists an $a_{0}$ such that the set $M$ is infinite. We suppose further that $a_{0}$ is minimal. Clearly $a_{0} \neq 0$, for if it were, the set $M = \{ m \in \mathbb{N} : m \leq 0\}$ = $\{0\}$ is finite.

Consider the set $N = M - \{n \in \mathbb{N}: a_{1} < n \leq a_{0}\}$, which is just the set $M$ with the finitely many naturals deleted. From elementrary set theory, $N$ must be infinite.

But $N = \{ n \in \mathbb{N} : n \leq a_{1}\}$, and we can suppose that $a_{1}$ is the first term in a natural number sequence in infinite descent. Thus we have found another infinite set with the desired property, but with $a_{1} < a_{0}$. This contradicts the minimality of $a_{0}$.