Deriving the Poisson Distribution (Limits)

[AvgStudent]

Background:
The binomial distribution requires us to know the number of successes x in n repeated trials. We assume the probability of success p is constant over each trial.
However, if we are given only one piece of information, the rate, [imath]\lambda[/imath] (the average # of events or the expected value), and we don't know anything about probability p, nor the number of trials n. The binomial distribution can no longer do the job.

Using the limit, the unit times are now infinitesimal. We no longer have to worry about more than one event occurring within the same unit time.
Mathematically, when n-> [imath]\infty[/imath], and p → 0, [math]p=\frac{\lambda}{n}[/math]
Show that [math]Pr(X=k)=\lim_{n \to \infty} {n \choose k}p^k(1-p)^{n-k}=\frac{e^{-\lambda}\lambda^k}{k!}[/math]My attempt:
[math]Pr(X=k)=\lim_{n \to \infty} \frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n-k}[/math][math]Pr(X=k)=\lim_{n \to \infty} \frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n}\left(1-\frac{\lambda}{n}\right)^{-k}=\lim_{n \to \infty} \frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^ke^{-\lambda}(1)[/math]I'm stuck and not sure how to continue.

 

[AvgStudent]

I've noticed that:
[math]\lim_{n \to \infty} \frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^ke^{-\lambda}(1)=\lim_{n \to \infty}\frac{n!}{(n-k)!(n^k)}\frac{e^{-\lambda}\lambda^k}{k!}[/math]So how to show that:
[math]\lim_{n \to \infty}\frac{n!}{(n-k)!(n^k)}=1?[/math]

 

[AvgStudent]

Is this a valid approach? [math]\lim_{n \to \infty}\frac{n!}{(n-k)!(n^k)}=\lim_{n \to \infty}\frac{n(n-1)(n-2)...(n-k+1)}{\underbrace{n*n*\dots *n}_{\text{k times}}}\\ =\lim_{n \to \infty}\left(\frac{n}{n}\right)\lim_{n \to \infty}\left(\frac{n-1}{n}\right)\dots \lim_{n \to \infty}\left(\frac{n-k+1}{n}\right)\stackrel{l'Hospital}{=}(1)(1)(1)\dots(1)=1^k=1[/math]

 

[BigBeachBanana]

looks good to me.

 

[AvgStudent]

Thanks, both.