Simple upper bound for $\binom{n}{k}$
I assume you are looking for the simple bound $$\binom{n}{k} < \left(\dfrac{e n}{k}\right)^k$$ Proof: \begin{align} \binom{n}{k} &= \frac{n(n-1)\dots(n-k+1)}{k!} \\ &= 1\left(1-\frac{1}n \right)\cdots\left(1-\frac{k-1}n \right) \frac{n^k}{k!}\\ &< \frac{n^k}{k!} \qquad \text{as all factors on the left are }\le 1. \end{align}
From the Taylor Series of $e^x$, we know $\forall k \in \mathbb{N}, \; e^k > \dfrac{k^k}{k!}$. Combining this with the above, we get the desired bound.