Simple upper bound for $\binom{n}{k}$

I remember seeing an upper bound for the binomial $\binom{n}{k}$ with an exponential function, something like $\binom{n}{k}\leq \left(ne/k\right)^k$. What exactly is it, and are there other similar good upper bounds for $\binom{n}{k}$?

Edit: As the link in Macavity’s comment shows, the bound is indeed $\binom{n}{k}\leq \left(ne/k\right)^k$. How can we prove this?

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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.