Please prove that ${n\choose r} < (n+1)^r$. My induction-based proof is ugly! I came across this inequality in the University of Missouri Youtube Channel lecture 38:
College Algebra - Lecture 38 - The Binominal Theorem
The professor asks us 1:01:00 into the video to prove that "$1001^{999} < 1000^{1000}$" and throws in the above inequality as a fact that we can use in our proof. Proving "$1001^{999} < 1000^{1000}$" using the assumption that ${n\choose r} < (n+1)^r$ is rather easy. But I don't want to accept it by rote that ${n\choose r} < (n+1)^r$. So I tried to prove it using induction for any value of $r$. I proved that the base case when $r=1$ is true. Then went on to prove that if it is true for any $r$, then it is true for $r+1$ as well. Although I proved it using induction, it's a really ugly and messy proof. So please give me a neat proof for this inequality. I am sure it would be very easy for any regular contributor in this forum. Thanks.
PS: Please bear with my wordiness at the moment. I just don't know yet how to format my text or use mathematical symbols on this forum (It will be nice of you if you can tell in passing where to learn the markup and formatting code for this forum).

Note that 
$$\binom{n}{r}=\frac{n!}{(n-r)!r!}=\prod_{k=1}^{r} \frac{n-(r-k)}{k}.$$
So we have that 
$$\binom{n}{r} \leq n\cdot (n-1)\cdots (n-r+1) < (n+1)^r.$$
