Why does $\binom{n}{k} = 0$, if $k > n$?

Question

I have came across this in a textbook that I am currently studying, but I don't understand how I should proof this.

A short explanation or proof would be nice.

Answer 1

Since $$ \binom{n}{k}=\prod_{j=0}^{k-1}\frac{n-j}{j+1}\tag1 $$ if $k\gt n\ge0$ and $n\in\mathbb{Z}$, then the product contains a factor of $0$.

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In the Generalized Binomial Theorem, $(1)$ is the definition of the binomial coefficients (with the convention that when $k=0$, $(1)$ is an empty product, and therefore, $1$). They are only defined for $k\in\mathbb{Z}$ and for $k\lt0$, $\binom{n}{k}=0$.

However, if $n\not\in\mathbb{Z}$, then $\binom{n}{k}\ne0$ for any integer $k\ge0$.

Answer 2

For example expand the binomial $$ (1+x)^n $$ and the coefficient of $x^k$ is $\binom{n}{k}$ for all $k$. [This is why it is called a binomial coefficient.] Of course (when $n$ is a positive integer), this coefficient is $0$ for $k > n$.