= Solution
Give the smooth curve its complex orientation. The <cellular homology of complex projective space> gives $H_2(\mathbb{CP}^2;\mathbb Z)=\mathbb Z$, generated by the class $[L]$ of a projective line. Thus $[C]=D[L]$ for a unique integer $D$, its <homological degree of a smooth complex plane curve>. Two distinct projective lines meet transversely in one point with positive sign, so $[L]\cdot[L]=1$. Therefore the degree $D$ is the <intersection number> $[C]\cdot[L]$ with any transverse projective line.
For the displayed curve, take $d\ge1$ and put $F=x^d+y^d-z^d$. If $d\ge2$, its three partial derivatives cannot vanish simultaneously at a projective point; if $d=1$, the equation defines a line and is also smooth. Intersect with $L=\{y=0\}$. Every intersection has $z\ne0$, so in the chart $z=1$ the points are
$$
[\zeta:0:1],\qquad\zeta^d=1.
$$
There are exactly $d$ distinct points. At each, $\partial F/\partial x=d\zeta^{d-1}\ne0$. In affine coordinates the curve is a graph $x=x(y)$, whereas the line is $y=0$; their tangent lines are transverse. For $d\ge2$ the curve tangent at the point is the $y$-direction, while for $d=1$ it is the direction $dx+dy=0$, still transverse to $dy=0$.
Each intersection contributes $+1$, not merely an unsigned count: both tangent spaces are complex lines, and the real <determinant> of the complex-linear isomorphism from their direct sum to the ambient tangent space is the squared modulus of its complex <determinant>, hence positive. Therefore $[C]\cdot[L]=d$, and
$$
\boxed{[C]=d[\mathbb{CP}^1],\qquad\deg C=d.}
$$
This computes the degree by explicit transverse intersections, without using Bézout's theorem to assume the answer.
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