The plane branch y²=x⁵ has Puiseux parameter (t²,t⁵)

$f$ is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$. By [F1] and [F2] it suffices to show that the monic quadratic $f=y^2-x^5\in {\mathcal{O}}_{\mathbb{C},0}[y]$ is irreducible. Suppose $f=GH$ with $G,H$ nonunits. Their leading coefficients multiply to the leading coefficient $1$, so both are units by [F3]. Neither factor can have degree $0$, since a degree-zero factor is its leading coefficient and would be a unit. Since their degrees sum to $2$, both have degree $1$; rescaling by their unit leading coefficients, we may write $G=y-a$, $H=y-b$ with $a,b\in {\mathcal{O}}_{\mathbb{C},0}$. Comparing coefficients gives $a+b=0$ and $ab=-x^5$, hence $a^2=x^5$, contradicting [F4]. Thus $f$ is irreducible in the polynomial ring and, by [F2], in the germ ring; by [F5] $f$ is reduced and $I_0(X)=(f)$.

$\gamma (t)=(t^2,t^5)$ is injective. Indeed $\gamma (t_1)=\gamma (t_2)$ means $t_1^2=t_2^2$ and $t_1^5=t_2^5$. The first equation gives $t_2=\pm t_1$; if $t_2=-t_1$, then $t_1^5=t_2^5=-t_1^5$, so $2t_1^5=0$ and $t_1=0=t_2$; otherwise $t_2=t_1$.

The image of $\gamma$ on ${\mathrm{\Delta }}_{\delta }=\{\mathrm{\mid }t\mathrm{\mid }<\delta \}$ is exactly the full representative $Z(f)\cap \{\mathrm{\mid }x\mathrm{\mid }<{\delta }^2\}$ of the germ $X$. First, $f(\gamma (t))=t^{10}-t^{10}=0$, so the image lies in $Z(f)$, and $\mathrm{\mid }{t}^2\mathrm{\mid }=\mathrm{\mid }t{\mathrm{\mid }}^2<{\delta }^2$. Conversely, let $(x,y)\in Z(f)$ with $\mathrm{\mid }x\mathrm{\mid }<{\delta }^2$; choose $t$ with $t^2=x$, so $\mathrm{\mid }t\mathrm{\mid }<\delta$. If $x=0$, then $y^2=0$ and $y=0=\gamma (0)$. If $x\mathrm{\ne }0$, then $y^2=x^5=x\cdot (x^2{)}^2$ gives $(y\mathrm{/}{x}^2{)}^2=x=t^2$, so $y\mathrm{/}{x}^2=\pm t$ and $y=\pm t^5$; replacing $t$ by $-t$ if necessary, we get $(x,y)=(t^2,t^5)=\gamma (t)$ with $t^2=x$. Hence every point of the representative is attained, and $h(t)=t^5={\sum }_{k>2}{a}_k{t}^k$ with $a_5=1$ has order $5>2$.

Every holomorphic parametrisation $s\mapsto (s^k,j(s))$ of the germ $X$ in these coordinates has $k\ge 2$. Such a parametrisation has image containing a full representative $X\cap U$ of the germ for some neighbourhood $U$ of $0$. Choose $\epsilon >0$ with $({\epsilon }^2,\pm {\epsilon }^5)\in U$; both points lie in $X$, because $(\pm {\epsilon }^5{)}^2={\epsilon }^{10}=({\epsilon }^2{)}^5$. So there are $s_1\mathrm{\ne }{s}_2$ with $s_i^k={\epsilon }^2$ and $j(s_1)={\epsilon }^5$, $j(s_2)=-{\epsilon }^5$; the two parameters are distinct since their images are. Thus the polynomial $T^k-{\epsilon }^2$ of degree $k$ has at least two distinct roots, so $k\ge 2$ by [F8].

$X$ is an irreducible hypersurface germ and its only singular point near $0$ is the origin. By step 1.1 the reduced defining germ $f$ is irreducible, so by [F5] the germ $X=Z(f)$ is irreducible: its decomposition has the single component $Z(f)=X$. The differential $df=(-5x^4,2y)$ vanishes at the origin and at no other point of $X$, because $x=0$ forces $y^2=x^5=0$ and then $y=0$. At a point $q\in X$ with $q\mathrm{\ne }0$ the translate of $f$ is a germ with nonzero differential, hence is not a product of two nonunits, that is, it is an irreducible and therefore reduced germ vanishing on $X$ near $q$; so it is a local reduced equation of $X$ at $q$ and [F6] makes $q$ a regular point.

Consequently $\gamma (t)=(t^2,t^5)$ is an injective convergent Puiseux parametrisation of $X$ in the standard coordinates: it is holomorphic on ${\mathrm{\Delta }}_{\delta }$, it is injective by step 1.2, the holomorphic function $h(t)=t^5$ satisfies $h(t)={\sum }_{k>2}{a}_k{t}^k$, and by step 1.3 its image germ is exactly $X=Z(f)$, matching the conclusion of [F7].

Steps 1.4, 2.1 and 3.1 prove all the assertions: $X$ is irreducible and singular only at the origin, $\gamma (t)=(t^2,t^5)$ is an injective convergent parametrisation of its germ, and since every parametrisation in the same coordinates has exponent $k\ge 2$ by step 1.4 while $\gamma$ has exponent $2$, the exponent is primitive (minimal). Over a base value $x_0\mathrm{\ne }0$ the two branch values are $\pm x_0^{5\mathrm{/}2}$, so the Puiseux exponent of this branch is $5\mathrm{/}2$, which differs from the exponent $3\mathrm{/}2$ of the cusp $y^2=x^3$.