We know that for a fixed $x>0$, the convergence of $$\left(1+\frac{x}{n}\right)^n\rightarrow e^x$$ is from below and so we have the following bound for any $n\in\mathbb{N}$:
$$\left(1+\frac{x}{n}\right)^n\leq e^{x}.$$
However, for a fixed $y<0$, $y=-x$ the convergence of $$\left(1+\frac{y}{n}\right)^n=\left(1-\frac{x}{n}\right)^n\rightarrow e^{-x}$$ is from above and so we certainly can't write $$\left(1-\frac{x}{n}\right)^n\leq e^{-x}.$$
EDIT: No it isn't and yes we can. See answer below.
Is there a well-known way to salvage an upper bound here? Perhaps a bound of the form $$\left(1-\frac{x}{n}\right)^n\leq \phi(x)e^{-x},$$ for some relatively small $\phi(x)$.
Context:
At the moment, for a fixed $n$, I am hoping to upper bound the term $$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k},$$
where I have control on $k$. I had mistakenly taken $k=\alpha(n^n-1)$ for some $\alpha>0$ and concluded that
$$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k}\leq e^{-2\alpha(n-1)(\sqrt{n}-1)^2},$$
but of course this is incorrect and I need something more along the lines of
$$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k}\leq \phi((n-1)(\sqrt{n}-1)^2)e^{-2\alpha(n-1)(\sqrt{n}-1)^2},$$
There is a misstatement in the OP. The sequence $e_n=\left(1-\frac xn\right)^n$ is actually an increasing sequence. To see this, we write for $x<n$
$$\begin{align} \frac{e_{n+1}}{e_n}&=\frac{\left(1-\frac x{n+1}\right)^{n+1}}{\left(1-\frac xn\right)^n}\\\\ &=\left(1-\frac xn\right)\left(1+\frac{x}{(n-x)(n+1)}\right)^{n+1} \tag 1\\\\ &\ge \left(1-\frac xn\right)\left(1+\frac{x}{(n-x)}\right) \tag 2\\\\ &=1 \end{align}$$
where in going from $(1)$ to $(2)$ we exploited Bernoulli's Inequality.
Therefore, $e_n$ increases monotonically for $x<n$ to $e^{-x}$ and
$$e^{-x}\ge \left(1-\frac xn\right)^n$$