= Solution
<Normalization of a probability distribution> and the <geometric series> give $1=A/(1-\rho)$, so $A=1-\rho$. For a nonnegative <integer> $n$, the strict tail starts at $n+1$:
$$
\boxed{\mathbb P(N>n)=\sum_{j=n+1}^\infty(1-\rho)\rho^j=\rho^{n+1}.}
$$
For nonnegative <integers> $n>m$, the <event> $\{N>n\}$ is contained in $\{N>m\}$. Therefore
$$
\boxed{\mathbb P(N>n\mid N>m)=\frac{\rho^{n+1}}{\rho^{m+1}}=\rho^{n-m}.}
$$
This is the <memorylessness of the geometric distribution> with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.
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