On finite endomorphisms of $\mathbf{P}^r$ This question is basically on applying the Grothendieck-Riemann-Roch theorem to finding a formula for the push-forward of a line bundle on $\mathbf{P}^r$ under a certain morphism. Since I have a lot of questions, let me begin with an example.
Let $\pi:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be the morphism defined as $[x_0:x_1] \mapsto [x_0^n:x_1^n]$. Here $\mathbf{P}^1$ denotes the projective line over $\mathbf{C}$ and $n\geq 1$ is an integer. Note that $\pi$ is finite. For, it is locally given by the map $x\mapsto x^n$.
Let $X=\mathbf{P}^1$.
One can show that $\pi_\ast (\mathcal{O}(m))$ is given (up to isomorphism) by $$ \mathcal{O}(n(m+1)-1)\oplus \ldots \oplus \mathcal{O}(n(m+1)-1).$$ Here the sum is taken $n$ times. For example, $\pi_\ast \mathcal{O}_X \cong \mathcal{O}(n-1)^{\oplus n}$.
This is wrong. See David Speyer's response for the correct expression.
Now for the questions.
Q1. I would like to look at the morphism $\mathbf{P}^r\longrightarrow \mathbf{P}^r$ given by $[x_0:\ldots:x_r]\mapsto [x_0^n:\ldots:x_0^r]$. This is a finite morphism and therefore it should be possible to apply GRR in finding an expression for $\pi_\ast \mathcal{O}(m)$, right? Now, this is probably still very easy so I was wondering if there were any other results in this direction.
Q2. How would one do this without the Grothendieck-Riemann-Roch theorem? By "this", I also mean the above example for $\mathbf{P}^1$.
Q3. What is the geometric interpretation of this? 
Final note. Since every vector bundle on $\mathbf{P}^1$ has a unique decomposition into twisted sheaves, we get a nice expression for $\pi_\ast \mathcal{E}$ where $\mathcal{E}$ is a vector bundle on $\mathbf{P}^1$.
 A: The claimed computation is still wrong. Let $m \equiv r \mod n$, with $0 \leq r < n$. Then the right answer is that 
$$\pi_* \mathcal{O}(m) = \mathcal{O}( \lfloor (m+1)/n \rfloor-1)^{\oplus(n-r-1)} \oplus \mathcal{O}( \lceil (m+1)/n \rceil-1)^{\oplus(r+1)}.$$
Let $S$ be the source $\mathbb{P}^1$ and $T$ the target. As a general piece of advice, you should never identify two spaces when you don't have to. Here are three ways you could get this answer:
 By Grothendieck-Riemmann-Roch
Let $L_S$ be the line bundle $\mathcal{O}(1)$ on $S$ and $L_T$ the line bundle $\mathcal{O}(1)$ on $T$. Let $H_S$ and $H_T$ be the hyperplane classes in $H^*(S)$ and $H^*(T)$. The chern character of $L_S^{\otimes m}$ is $(1+H_S)^m = 1+m H_S$. The Todd classes of $S$ and $T$ are $1+H_S$ and $1+H_T$. So $\pi_* L_S^{\otimes m}$ is something with chern character 
$$(1+H_T)^{-1} \pi_*\left( (1+m H_S) (1+H_S) \right) = (1+H_T)^{-1} \pi_*\left( 1+(m+1) H_S \right).$$
Note that $\pi_* 1 = n$ and $\pi_* H_S = H_T$. So we get
$$ch(\pi_* \mathcal{O}(m)) = (1+H_T)^{-1}  (n+(m+1) H_T)=n + (m-n+1) H_T.$$
(We know that $R^1 \pi_*$ vanishes because the map is finite.)
From the leading term, we see that $\pi_* \mathcal{O}(m)$ has rank $n$. It is not completely obvious that is torsion free. If we assume it is, then it must be of the form $\bigoplus \mathcal{O}(a_i)$ for some $a_1 + \cdots + a_n$. We see from the above computation that $\sum a_i = m-n+1$. 
That's as far as we can get from GRR. Grothendieck-Riemann-Roch can only do the computation in $K$-theory, so we can't distinguish $\mathcal{O}(-1) \oplus \mathcal{O}(1)$ from $\mathcal{O}(0) \oplus \mathcal{O}(0)$. 
Directly in K-theory
The point of Grothendieck-Riemann-Roch is that it gives a commuting diagram
$$\begin{matrix} K^0(S) & \longrightarrow & H^*(S) \\
\downarrow & & \downarrow \\
K^0(T) & \longrightarrow & H^*(T). \end{matrix}$$
I usually find that it is just as easy to do my computations directly on the left hand side of the diagram. Let $p_S$ and $p_T$ be the class of the structure sheaf of a point on $S$ and $T$. We have the short exact sequence $0 \to \mathcal{O}(-1) \to \mathcal{O}(0) \to \mathcal{O}_{\mathrm{pt}} \to 0$, so $p_S = 1-L_S^{-1}$ and $L_S = 1+p_S$. (Since $p_S^2=0$.) 
Clearly, $\pi_* p_S = p_T$. So 
$$\pi_* \mathcal{O}(m) = \pi_* (1+p_S)^m = \pi_* (1+m p_S) = \pi_* 1 + m p_T.$$
We can see that $\pi_* 1$ has rank $n$; say $\pi_* 1 = n+a p_T$.
Let $\chi$ be pushforward to the $K$-theory of a point, better known as holomorphic Euler characteristic. 
Since pushforward is functorial, the sequence of maps $S \to T \to \mathrm{pt}$ shows that $$\chi(\pi_* 1) = \chi(1) = 1$$
so $n+a=1$ and $a = -(n-1)$. We see that
$$\pi_* \mathcal{O}(m) = n + (m-n+1) p_T.$$
By direct computation
It is easy enough to do this example, and any toric example like it, directly from the definition of pushforward. As a bonus, this will tell us exactly which vector bundle we get, not just the $K$-class.
Let $S_1 \cup S_2$ be the open cover of $S$ where $S_1 = \mathrm{Spec} \ k[s]$ and $S_2 = \mathrm{Spec} \ k[s^{-1}]$. Similarly, define $T_1$, $T_2$, $k[t]$ and $k[t^{-1}]$. 
Let $e$ be a generator for the free $k[s]$ module $\mathcal{O}(m)(S_1)$. Then $s^m e$ is a generator of $\mathcal{O}(m)(S_2)$. By definition, $\left( \pi_* \mathcal{O}(m) \right) (T_1)$ is $\mathcal{O}(m)(S_1)$ considered as a $k[t]$-module. As such, it has basis
$$ e,\ s e,\ s^2 e,\ \ldots s^{n-1} e. \quad (*)$$
Similarly,  $\left( \pi_* \mathcal{O}(m) \right) (T_2)$ has basis
$$ s^m e,\ s^{m-1} e,\ \ldots, s^{m-n+1} e. \quad (**)$$
Reorder the lists $(*)$ and $(**)$ so that corresponding elements have exponents of $s$ which are congruent modulo $n$. To keep notation simple, I'll do the case of $m=0$. So we pair off: 


*

*$e$ and $e$ 

*$s e$ and $s^{-n+1} e = t^{-1} (s e)$

*$s^2 e$ and $s^{-n+2} e = t^{-2} (s e)$
and so forth.


There is one time that we pair $v$ with itself and $(n-1)$ times that we pair $v$ with $t^{-1} v$. So the transition matrix between our bases is diagonal with entries $(1,t^{-1}, t^{-1}, \ldots, t^{-1})$ and the vector bundle is $\mathcal{O}(0) \oplus \mathcal{O}(-1)^{n-1}$. 
For general $m$, if I didn't make any errors, we get the formula at the beginning of the post.
