Paths that never go below the axis: Catalan numbers

Idea. The reflection principle: bad paths are in bijection with paths that are easy to count.

All paths. A path from $0$ to $0$ with $2n$ steps has $n$ up-steps and $n$ down-steps, so there are $\binom{2n}n$ of them.

Bad paths. A path from $0$ to $0$ is bad if it reaches height $-1$ at some moment. Consider the first such moment $t$, and reflect the portion of the path before $t$ about the line $y=-1$. The new path starts at $-2$, still reaches $-1$ first at time $t$, and ends at $0$.

Conversely, any path from $-2$ to $0$ with $2n$ steps must cross $-1$; reflecting the part before the first visit to $-1$ gives a bad path from $0$ to $0$. These operations are inverse, so bad paths are in bijection with all paths from $-2$ to $0$.

Counting those. A path from $-2$ to $0$ in $2n$ steps has $u$ up-steps and $d$ down-steps with $u+d=2n$ and $u-d=2$: $u=n+1$, $d=n-1$. So there are $\binom{2n}{n+1}$ of them.

Conclusion.
$$C_n=\binom{2n}n-\binom{2n}{n+1}=\frac1{n+1}\binom{2n}n.$$
The first values are $1,2,5,14,42,132,\dots$

Remark. The same numbers count well-formed sequences of $n$ pairs of brackets and triangulations of a convex $(n+2)$-gon.

Lesson. Reflection principle: reflect an initial segment to transform a constrained count into an unconstrained one.