I think I've finally managed to solve the following exercise from Hartshorne but I would really appreciate someone checking it out for me (specifically, I'm looking for someone to say "looks good" or "no it's wrong/is incomplete and here's why...").
Many thanks!
Let $\rho:\mathbb{P}^n\to \mathbb{P}^N$ be the Veronese embedding (where $N=\binom{n+d}{d}-1$).
Let $S$ be the set of multi-indices: $$\{(i_0,i_1,\ldots,i_n) \, : \, \text{each } i_j \in \{0,1,\ldots,d\} \text{ and } i_0+i_1+\ldots+i_n=d\}.$$ Let $\theta:k\big[\{y_I \, : \, I \in S\}\big] \to k[x_0,\ldots,x_n]$ be the $k$-algebra homomorphism such that $$\theta(y_I)=x_0^{i_0}x_1^{i_1}\ldots x_n^{i_n} \; \; \; \; \forall I \in S.$$
I want to show that $\mathrm{im}\,\rho=Z(\ker\theta)$ and that $\rho$ is a homeomorphism onto its image.
My proposed solution follows...
1) I show that $\rho$ maps $\mathbb{P}^n$ injectively into $Z(\ker\theta).$
For any $(a_0:a_1:\ldots:a_n) \in \mathbb{P}^n$ and any $f \in \ker(\theta),$ we have $$f(\rho(a_0:a_1:\ldots:a_n))=\theta(f)(a_0,a_1,\ldots,a_n)=0.$$ Moreover, if $\rho(a)=\rho(b)$ for some $a,b \in \mathbb{P}^n$ with $a_i\neq 0,$ then $$a=(a_0a_i^{d-1}:\ldots:a_i^d:\ldots:a_na_i^{d-1})=(b_0b_i^{d-1}:\ldots:b_i^d:\ldots:b_nb_i^{d-1})=b.$$ Thus $\rho$ realises $\mathbb{P}^n$ as a subset of $Z(\ker\theta).$
2) I construct local inverses to $\rho.$
Observe that, for each $(i_0,i_1,\ldots,i_n) \in S,$ we have the relation: $$d\cdot(i_0,i_1,\ldots,i_n)=i_0\cdot(d,0,\ldots,0)+\ldots+i_n\cdot(0,\ldots,0,d).$$ It follows from this that the $n+1$ affine-open patches $U_{(d,0,\ldots,0)},\ldots,U_{(0,\ldots,0,d)}$ cover $Z(\ker\theta).$
Define a map $\sigma_0:U_{(d,0,\ldots,0)}\cap Z(\ker\theta)\longrightarrow U_0\subset \mathbb{P}^n$ as follows $$\sigma_0(b)=(b_{(d,0,\ldots,0)}:b_{(d-1,1,0,\ldots,0)}:\ldots:b_{(d-1,0,\ldots,0,1)}).$$ For each $(i_0,i_1,\ldots,i_n) \in S,$ we have the relation: \begin{align*}i_0\cdot(d,0,\ldots,0)+i_1\cdot(d-1,1,0,\ldots,0)+\ldots&+i_n\cdot(d-1,0,\ldots,0,1)\\[1em] &=(i_0,i_1,\ldots,i_n)+(d-1)\cdot(d,0,\ldots,0).\end{align*} It follows from this that $(\rho\circ\sigma_0)(b)=b$ for all $b \in U_{(d,0,\ldots,0)}\cap Z(\ker\theta).$
Furthermore, it is clear that $(\sigma_0\circ\rho)(a)=a$ for all $a \in U_0,$ and so $\sigma_0$ and $\rho|_{U_0}$ are mutually inverse bijections.
In a similar fashion, for each $i=1,\ldots,n,$ we construct a map $\sigma_i:U_{(0,\ldots,0,d,0,\ldots,0)}\cap Z(\ker\theta) \longrightarrow U_j$ inverse to $\rho|_{U_i}.$
3) I show that the $\sigma_i$ can be glued together to form a global inverse for $\rho.$
If $b \in \mathrm{dom}(\sigma_i)\cap\mathrm{dom}(\sigma_j)$ for $i\neq j,$ then $\rho(\sigma_i(b))=b=\rho(\sigma_j(b))$ and so $\sigma_i(b)=\sigma_j(b)$ since $\rho$ is injective by (1).
Therefore the $\sigma_i$ agree on all overlaps and we patch them together to get $\sigma:Z(\ker\theta) \longrightarrow \mathbb{P}^n,$ a global inverse to $\rho.$
4) I explain why $\rho$ is a homeomorphism onto its image.
Since both $\rho$ and $\sigma$ are defined (locally, in $\sigma$'s case) by homogeneous polynomials, they are both continuous.
It thus follows that $\rho:\mathbb{P}^n \to Z(\ker\theta)$ is a homeomorphism.