How to prove this property in a Poisson process? For a Poisson process show, for $s < t$, that
$P(N(s)=k|N(t)=n) = \binom{n}{k} (\frac{s}{t})^k (1-\frac{s}{t})^{n-k}$

The joint probability mass function for $(N(s), N(t))$ for $s<t$ is, using the property of independent increments, i.e. $N(t) \sim N(s) + N(t-s)$:
$$
  \mathbb{P}(N(s) = k, N(t) = n) = \mathbb{P}(N(s) = k, N(t) - N(s) = n - k) = \\\frac{(\mu s)^k}{k!} \mathrm{e}^{-\mu s} \cdot \frac{(\mu (t-s))^{n-k}}{(n-k)!} \mathrm{e}^{-\mu(t-s)} \cdot [ n-k \geqslant 0, k \geqslant 0] 
$$
Then, assuming $n \geqslant k \geqslant 0$:
$$
  \mathbb{P}(N(s) =k | N(t) = n) = \frac{\mathbb{P}(N(s)=k,N(t)=n)}{\mathbb{P}(N(t)=n)} = \\
     \frac{\frac{\mu^k s^k}{k!} \cdot \frac{\mu^{n-k} (t-s)^{n-k}}{(n-k)!} \cdot \mathrm{e}^{-\mu t} }{ \frac{\mu^n t^n}{n!} \mathrm{e}^{-\mu t}} = 
    \left( \frac{s}{t} \right)^k \left( 1- \frac{s}{t} \right)^{n-k} \binom{n}{k}
$$
