Expectation of geometric random variable Let $X$ be geometric random variable with parameter $p$.
How to prove that:

(1) $E[X-1|X>1] = E[X]$
(2) $E[X^2|X>1] = E[(X+1)^2]$

Author explains the fact below and states it is used to prove (1)
$$P(X-1=k|X>k) = P(X=k).$$
I understood how the fact is true but could not understand how it is used to derive (1)
(2) was stated without explanation. Could someone help in deriving (1) and (2)?
 A: Firstly, I think your equation $P(X-1=k\mid X\gt k) = P(X=k)$ is supposed to be (proof below): $$P(X-1=k\mid X\gt 1) = P(X=k).\qquad\qquad (*)$$
(1)
\begin{eqnarray*}
E(X-1\mid X\gt 1) &=& \sum_{k=1}^{\infty}{kP(X-1=k\mid X\gt 1)} \\
&=& \sum_{k=1}^{\infty}{kP(X=k)} \qquad\qquad\text{using $(*)$} \\
&=& E(X).
\end{eqnarray*}
(2)
\begin{eqnarray*}
E(X^2\mid X\gt 1) &=& \sum_{k=1}^{\infty}{k^2P(X=k\mid X\gt 1)} \\
&=& \sum_{k=1}^{\infty}{k^2P(X=k-1)} \qquad\qquad\text{using $(*)$ with $k-1$ instead of $k$} \\
&=& \sum_{k=1}^{\infty}{(k+1)^2P(X=k)} \qquad\qquad\text{re-index the sum using $P(X=0)=0$} \\
&=& E((X+1)^2).
\end{eqnarray*}
Note the different mechanism in first line of the two methods. In (1) we treat $X-1$ as a random variable in itself, whereas in (2) we treat $X^2$ as a function of the random variable $X$.
$\\$
Claim: $P(X-1=k\mid X\gt 1) = P(X=k)$.
Firstly, for $k=0$ it's clear that both sides are $0$. So now assume $k\gt 0$.
\begin{eqnarray*}
P(X-1=k\mid X\gt 1) &=& P(X-1=k\cap X\gt 1) / P(X\gt 1) \\
&=& P(X=k+1\cap X\gt 1) / P(X\gt 1) \\
&=& P(X=k+1) / (1-P(X=1)) \\
&=& (1-p)^kp / (1-p) \\
&=& (1-p)^{k-1}p \\
&=& P(X=k).
\end{eqnarray*}
