Binomial CDF - Wikipedia vs Reality The Binomial distribution article on Wikipedia defines the binomial CDF as 
$F_{Bin(n,p)}(k) = I_{1-p}(n-k, k+1)$,
where $I$ is the regularized incomplete beta function.
There is a proof for the equality on Mathematics Stackexchange, but I stumbled into a little problem. When $n=k$ we get that $F_{Bin(n,p)}(n) = I_{1-p}(0, n+1)$, but $I$ is not defined for $0$.
I am sure I don't understand this correctly, but I can't understand where I'm wrong.
Edit: My friends and I thought about this definition: $F_{Bin(n,p)}(k)=lim_{m \rightarrow k} I_{1−p}(n−m,m+1)$. I think it works, but did not find any citation to support it.
 A: Whuber's link explains (ab initio) the Relationship between Binomial and Beta distributions 
The binomial distribution, which can be seen as a sum of $n$ Bernoulli experiments $y_i$, is equivalent to picking $n$ variables $x_i$ from a uniform distribution and assigning $y_i = 1$ if $x_i \leq p$ and $y_i = 0$ if $x_i > p$. 
Then the probability that the sum of Bernouilli experiments $z=\sum y_i$ is equal or above some number $k$ relates to the probability that the $k$-th order variable in the uniform variables is less than $p$ (because $Z \geq k$ if and only if $X_{(k)} \leq p$). 
Thus:
$$P(Z \geq k) \equiv P(X_{(k)} \leq p)$$ 
where $$X_{(k)} \sim \text{Beta}(k,n+1-k)$$
and
$$P(X_{(k)}\leq p) =I_{p}(k,n+1-k) = \frac{\int_0^p x^{k-1} (1-x)^{n-k} dx}{ \int_0^1 x^{k-1} (1-x)^{n-k}dx}$$

So it is easy to express with an intuitive analogy.
But, when $k = 0$ (which relates to your expression $k=n$) then you are looking for $P(X_{(0)}\leq p)$, this has no meaning although you could assign a value 1 to it if you like. 
For this case the analogy and how you build up the expression for the CDF makes no sense. So, indeed, the expression from Wikipedia, with the Beta distribution, does not work (because the analogy of drawing from uniform distribution is limited to ordered variables $1$ to $n$, which excludes a zero).
If you use 
$$P(Z \leq k-1) \equiv 1- P(Z \geq k)$$ 
then it doesn't make sense to use $k=0$ anyway. Although you get back the problem at the other end that you will be looking for $k=n+1$ and the meaningless $P(Z \geq n+1)$ and $P(X_{(n+1)} \leq p)$ (if you like you could say those probabilities are 0, although the expression with the beta function breaks down). 
