prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$ How to prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$. 
By memoryless property, 
$$P(X=i+s | X>i)=P(X=s)$$
How to get distribution for X from the above ?
 A: Let $p=P[X\geqslant2]$, then $P[X\geqslant i+1\mid X\geqslant i]=p$ for every $i\geqslant1$ hence $P[X\geqslant i]=p^{i-1}$ for every $i\geqslant1$. Surely you can deduce the distribution of $X$ from this observation.
A: First, observe that
$$
\mathbb{P}\{X>t\}=\sum_{n > t}\mathbb{P}\{X=n\}
$$
so one can show that the memorylessness property implies ($s,t>0$)
$$
\mathbb{P}[X>s+t| X > t]=\mathbb{P}\{X> s\}
$$
Note also that since $s>0$
$$
\mathbb{P}[X>s+t| X > t\} = \frac{\mathbb{P}\{X>s+t\}}{\mathbb{P}\{X > t\}}
$$
Define $f\colon t\in\mathbb{R}_+\mapsto\mathbb{P}\{X>t\}\in[0,1]$. We have $f\searrow$, and furthermore $\forall t,s\geq 0$
$$
 \frac{f(t+s)}{f(t)}=f(s)
$$
i.e
$$
 f(s)f(t)=f(s+t)\qquad \forall s,t \geq0
$$
Now, remains to solve this functional equation (with additional condition that $f$ is non-increasing) to get the only possible form(s) for $\mathbb{P}\{X> t\}$ — which characterizes $X$'s law.
A: This is essentially a reworking of Did's answer. Fix $i=1$, $q = P\{X > 1\}$
and note that
$$P\{X = s+1\mid X > 1\} = \frac{P\{X = s+1, X > 1\}}{P\{X > 1\}}
  = \frac{P\{X = s+1\}}{q} = P\{X=s\}$$
which gives
$$\begin{align}
P\{X=2\} &= qP\{X=1\}\\
P\{X=3\} &=q P\{X=2\} = q^2P\{X=1\}\\
P\{X=4\} &=q P\{X=3\} = q^3P\{X=1\}\\
\vdots\qquad &\vdots \qquad \vdots\\
P\{X = n\} &= qP\{X = n-1\} = q^nP\{X=1\}\\
\vdots\qquad &\vdots \qquad \vdots
\end{align}$$
so the probability masses decrease as a geometric series.
