If $\{p_0,\dots,p_m\}$ is affine independent with barycentre $b$, then $\{b,p_0,\dots,\hat{p}_i,\dots,p_m\}$ is affine independent for each $i$. My question is how to solve this problem. I feel like I am missing something obvious here. (This is Exercise 2.9 in Rotman's Introduction to Algebraic Topology.)

If $\{p_0,\dots,p_m\} \subset \mathbf{R}^n$ is affine independent with barycentre $b$, then $\{b,p_0,\dots,\hat{p}_i,\dots,p_m\}$ is affine independent for each $i$. (By the way, the circumflex means 'delete'.)

Proof: See below. 
 A: 
If $\{p_0,\dots,p_m\} \subset \mathbf{R}^n$ is affine independent with barycentre $b$, then $\{b,p_0,\dots,\hat{p}_i,\dots,p_m\}$ is affine independent for each $i$. (By the way, the circumflex means 'delete'.)

Proof: Since affine independence does not depend on the ordering, we may assume that we have removed $p_0$.
We want to show that $\{b, p_1,\dots,p_m\}$ is affine independent where 
$$b = \frac{1}{m+1} (p_0 + \cdots + p_m).$$ 
To do this, it is enough to show that the set $\{b-p_1,\dots,b-p_m\}$ is a linearly independent subset of $\mathbf{R}^n$ (when viewed as a vector space over $\mathbf{R}$). Note that for each $i$, we have 
$$b - p_i = \frac{1}{m+1} (p_0 + \cdots + \hat{p}_i + \cdots + p_m) - \frac{m}{m+1}p_i.\tag{1}$$ 
Suppose that $\{s_1, \dots, s_m\} \subset \mathbf{R}$ are scalars such that 
$$\sum_{i=1}^m s_i(b-p_i) = 0.\tag{2}$$
We want to show that $s_1=\cdots=s_m=0$.
Using (1) and (2) we can write
$$0 = \left(\frac{1}{m+1}\right)\sum_{i=1}^m ( s_i((p_0 + \cdots + \hat{p}_i + \cdots + p_m) - m p_i)),$$
and so
$$0 = \sum_{i=1}^m s_i(p_0 + \cdots - mp_i + \cdots + p_m),$$
whence
$$0 = - m \sum_{i=1}^m s_i p_i + p_0 \sum_{i=1}^m s_i + p_1 \sum_{i=1\\i\neq 1}^m s_i + \cdots + p_k \sum_{i=1\\i\neq k}^m s_i + \cdots + p_m \sum_{i=1\\i\neq m}^m s_i,$$
and finally
$$m \sum_{i=1}^m s_i p_i = p_0 \sum_{i=1}^m s_i + p_1 \sum_{i=1\\i\neq 1}^m s_i + \cdots + p_k \sum_{i=1\\i\neq k}^m s_i + \cdots + p_m \sum_{i=1\\i\neq m}^m s_i.$$
For the next bit of the proof we 'compare coefficients' of the $p_i$. 
For $p_0$ we have
$$0 = \sum_{i=1}^m s_i \implies s_1 = - \sum_{i=2}^m s_i.$$
Seeking a contradiction, suppose that $s_1 \neq 0$. Consider the coefficients of $p_1$. We have
$$ms_1 = \sum_{i=2}^m s_i \implies m\left(-\sum_{i=2}^m s_i\right) = \sum_{i=2}^m s_i.$$
The assumption that $\sum_{i=2}^m s_i \neq 0$ means that we can cancel on both sides, obtaining $m = -1$. Absurd!
So it must be the case that $s_1 = 0$.
It is easy to see how this can be iterated to show that the rest of the $s_i$ are all zero. So we'll stop writing here.
