Why is $(T-\lambda I)^{p-1}(x)$ an eigenvector? 
Definition. Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $\lambda$ if $(T-\lambda I)^p(x)=0$ for some positive integer $p$.
Notice that if $x$ is a generalized eigenvector of $T$ corresponding to $\lambda$, and $p$ is the smallest positive integer for which $(T-\lambda  I)^p(x)=0$, then $(T-\lambda I)^{p-1}(x)$ is an eigenvector of T corresponding to $\lambda$. Therefore $\lambda$ is an eigenvalue of $T$.

Are we saying that the smallest positive integer $p$ is $1$, and so $p-1=0$, and then $(T-\lambda  I)^{p-1}(x)=(T-\lambda I)^0(x)=x \neq 0$?
 A: We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $\lambda$ then for some $n$ we have $(T-\lambda I)^n(x) =0$. 
Note that this $n$ is not unique, because $(T-\lambda I)^{n+1}(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).
Then if we set $v = (T-\lambda I)^{p-1}(x)$ then $v\neq 0$ by minimality of $p$, and we have $(T-\lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $\lambda$. 
A: You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.
You have $(T-\lambda I)^p(x)=(T-\lambda I)((T-\lambda I)^{p-1}(x))=\mathbf 0$, i.e. per definition $(T-\lambda I)^{p-1}(x)\in\mathrm{ker}(T-\lambda I)$, i.e. $(T-\lambda I)^{p-1}(x)$ is an eigenvector.
Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-\lambda I)^p(x)=\mathbf0$, i.e. $(T-\lambda I)^{p-1}(x)\neq \mathbf0$. Note, that if $(T-\lambda I)^p(x)=\mathbf0$, then $(T-\lambda I)^{p+k}(x)=\mathbf0$ for any $k$ in the similar way as above.
A: The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-\lambda I)^{p-1}(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-\lambda I)^{p-1}(x) \neq 0$ for the reason you have stated (for $p = 1$, $(T-\lambda I)^{p-1}(x) = x \neq 0$ by assumption).

In general, suppose $p$ is the least positive integer such that $(T-\lambda I)^p(x) = 0$. Consider the vector $y = (T-\lambda I)^{p-1}(x)$. We know that $y \neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-\lambda I)^{p-1}(x) = 0$, which contradicts that $p$ is the least such positive integer.
Now, we write $$(T-\lambda I)^{p}(x) = [(T-\lambda I) \circ (T-\lambda I)^{p-1}](x) = (T-\lambda I)[(T-\lambda I)^{p-1}(x)] = (T-\lambda)(y).$$ 
Since $(T-\lambda I)^{p}(x)=0$ by assumption, we have that
$$
0 = (T-\lambda I)(y) \implies 0 = Ty - \lambda y \implies Ty = \lambda y.
$$
This shows that $y=(T-\lambda I)^{p-1}(x)$ is an eigenvector of $T$ with eigenvalue $\lambda$ when $p$ is the smallest positive integer such that $(T-\lambda I)^p(x) = 0$.
