# Step in proof of poisson probability

for the case when $$s < t$$

Let $$N(t)$$ be the amount of arrivals occurring at time period t in a Poisson process.

When $$N(t)$$ is known, the number of events by time $$s$$ can then be shown to follow a binomial distribution with parameters $$(N(t), \frac{s}{t})$$ as follows:

$$P(N(s) = k|N(t) = n) = \frac{p(N(s) = k, N(t) = n)}{P(N(t) = n)}$$ =

$$\frac{P(N(s) = k, N(t) - N(s) = n-k)}{P(N(t) = n)}$$ =

$$\frac{\frac{(\lambda s)^{k}}{k!}e^{-\lambda s} \cdot \frac{(\lambda(t-s))^{n-k}}{(n-k)!} e^{-\lambda(t-s)}}{\frac{(\lambda t)^{n}}{n!}e^{-\lambda t}}$$ =

$$\frac{s^{k}(t - s)^{n-k}}{t^{n}} \cdot \frac{n!}{k!(n-k)!} =$$

$$\binom{n}{k} (\frac{s}{t})^{k} (1 - \frac{s}{t})^{n-k}$$

I understand all of the above-mentioned steps accept part of the last one.

Particularly, why does $$\frac{s^{k}(t - s)^{n-k}}{t^{n}} = (\frac{s}{t})^{k} (1 - \frac{s}{t})^{n-k}$$?

$$\bigg(\frac{s}{t}\bigg)^k\bigg(1-\frac{s}{t}\bigg)^{n-k}$$ $$=\bigg(\frac{1}{t}s\bigg)^k\bigg[\frac{1}{t}(t-s)\bigg]^{n-k}$$ $$=\bigg(\frac{1}{t}\bigg)^ks^k\bigg[\frac{1}{t}\bigg]^{n-k}(t-s)^{n-k}$$ $$=\bigg(\frac{1}{t}\bigg)^ns^k(t-s)^{n-k}$$ $$=\bigg(\frac{1}{t^n}\bigg)s^k(t-s)^{n-k}$$ $$=\frac{s^k(t-s)^{n-k}}{t^n}$$