I was recently introduced to the set of Catalan numbers, which are of the form $\prod_{i = 2}^n \frac{n + i}{n}$. The first few Catalan numbers (which I might as well mention are all integers) are: 1, 1, 2, 5, 14, 42, 132, 429, 1430.
I've recently learned that Catalan numbers represent the number of paths that, in creating paths of lattice points from the origin to $(n, n)$, only intersect the line $y = x$ at the points $(0, 0)$ and $(n, n)$, where $n \in \mathbb{Z}$.
I've also learned that the number of such paths from $(0, 0)$ to $(n, n)$ that intersect $y = x$ at the points $(0, 0)$, $(n, n)$, and exactly one other point in between, is $C_{n-1}.$
How would I go about proving these two statements? They are undoubtedly related, in both their content and their similar structure. Any steps in the right direction would be appreciated.
Let me start with your second question, since this is probably going to be harder to Google.
The Catalan numbers are well-known to satisfy the following recurrence: $$ C_n = \sum_{k=1}^{n} C_{k-1}C_{n-k}$$ For instance, your first lattice path interpretation of the Catalan numbers $C_n$ satisfies this by conditioning on the first place after the origin that the paths touch the diagonal. You have to think about that for a minute, but the point is the $k-1$ shows up because to ensure that $k$ is the first time you hit the diagonal, you have to "step out one square" for the steps between the origin and $(k,k)$.
So you can see where that argument takes you. Let $D_n$ be the number of lattice paths that stay strictly above the diagonal except at $(0,0)$ and $(n,n)$ and exactly one other point. So if you want $k$ to be the only time you hit the diagonal, then you have to step out one square on both sides of $(k,k)$:
$$D_n=\sum_{k=1}^{n-1} C_{k-1}C_{n-k-1}$$
By comparing the two RHS, we can clearly see that $D_n=C_{n-1}$.
When you understand this argument, it's straightforward to unwind it into an explicit bijection between the $D_n$-type paths and the Catalan paths of length $2(n-1)$, but I'll leave that to you :)
Your first question has numerous answers on e.g. the Wikipedia page.
The recurrence above can be turned into the product formula that you mention; it is a standard exercise in generating functions and indeed is usually the first solution to a nonlinear recurrence relation that is solved in this way. You can find the details in any decent book on combinatorics, or the first proof on Wikipedia, or on this website: e.g. Simplifying Catalan number recurrence relation.
There are also direct proofs. They are all at least a little tricky, but you can get a sense of what they entail by looking at the other Wikipedia proofs. The way I understand this trickiness is that for your product formula to be an integer, it must be the case that $n+1$ is a divisor of $\binom{2n+1}{n}$. This in turn can be understood as the freeness of a natural cyclic group action; this and a number of similar/related phenomenon in algebraic combinatorics go by the name of "The Cycle Lemma".
In any case, there are many ways in which I think Catalan objects are more naturally quotients of the set of (appropriate) lattice paths, rather than subsets— and so I view the weirdness in these proofs as the same weirdness inherent anytime we work with "distinguished representatives" of equivalence classes, rather than the classes themselves.
(On a brief look I wasn't able to find direct proofs on MSE. Surely they're here; vets feel free to tell us where in the comments ^.^)