What's the nonstandard way to argue $\lim_{n\to\infty}\sum_{k=0}^n\binom n k (\frac x n)^k = \sum_{k=0}^\infty \frac{x^k}{k!}$ First, the equality holds, as:
$$\lim_{n\to\infty}\sum_{k=0}^n\binom n k \left(\frac x n\right)^k  =\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^n\right) = e^x =
 \sum_{k=0}^\infty \frac{x^k}{k!} $$
Having a limes over a series, this should be a prime target for nonstandard-reasoning, as we can just substitute in $h$ for $n$ where $h$ is an infinite hyperreal and drop the limit:
$$\sum_{k=0}^h\binom h k \left(\frac x h\right)^k \tag{i}$$ 
Now all that should be left to do should be to show that this sum is infinitesimally close to $\displaystyle  \sum_{k=0}^\infty \frac{x^k}{k!}$ 
So, by transforming $(i)$, we get $$\displaystyle\sum_{k=0}^h\binom h k \left(\frac x h\right)^k  = 
\sum_{k=0}^h \frac{h!}{k!(h-k)!} \left(\frac x h\right)^k
= 
\sum_{k=0}^h \frac{h!}{h^k(h-k)!} \frac {x^k} {k!} $$
Now, we split the sum in two parts: The natural summands, and the (infinite) hypernatural summands. 
 The natural summands should be infinitesimally close to our target sum , and the hypernatural summands should be infinitesimally close to 0.
Here's where my argumentation turns grotesque. As far as I know, comparing infinite hyperreals is a fruitless endeavor, as their difference/ratio can be any hyperreal value. So, I have to argue using the known limits of regular sequences:
$$ 
\lim_{k\to\infty} \frac {x^k}{k!} = 0  \qquad
\lim_{k\to\infty} \frac {n!}{n^k} = 0  
$$
Using these both limits, we can argue (Let $d_1,d_2$ be infinitesimals):
$$
\sum_{k\in\mathbb{^*N/N}\\k\le h}^h \frac{h!}{h^k(h-k)!} \frac {x^k} {k!} \approx 
\sum_{k\in\mathbb{^*N/N}\\k\le h}^h \frac{h!}{h^k(h-k)!} d_1 \approx
\sum_{k\in\mathbb{^*N/N}\\k\le h}^h \frac{d_1 d_2}{(h-k)!}\ \approx 0 $$
The argumentation here is that every summand $a_k$ is infinitely smaller than $h$, so even the infinite sum is infinitesimally close to $0$. (Question 1: Is this argumentation correct, and can it be made easier?)
And for the natural part of the sum, using $\frac {h-i}{h} \approx \frac {h}{h}\approx 1$ , for $i\in\mathbb{R}$:
$$
\sum_{k\in\mathbb{N}} \frac{h!}{h^k(h-k)!} \frac {x^k} {k!} =
\sum_{k\in\mathbb{N}} \frac{(h\cdot (h-1)\cdots(h-k+1))}{h^k} \frac {x^k} {k!}
\underbrace{\approx}_{(1)}
\sum_{k\in\mathbb{N}} \frac{h^k}{h^k} \frac {x^k} {k!} = 
\sum_{k\in\mathbb{N}} \frac {x^k} {k!}$$
Note that (1) holds as there's only finitely many $\approx$-transformations, i.e. the error introduced can't sum up to a significant amount.
(Question 2: Is this argumentation correct? It feels fishy)

Reacting on the spotted mistakes I've tried to fix the reasoning. Instead of splitting the sum $\displaystyle\sum_{k=0}^h$ in $\displaystyle\sum_{k\in\mathbb{N}}$ and $\displaystyle\sum_{k\in\mathbb{^*N/N}}$ I'm now trying a split into these two sums: 
$\displaystyle\sum_{k=0}^{h_1}$ and $\displaystyle\sum_{k=h_1}^{h_2}$ where $h_1,h_2$ are both infinite hyperreals. Now, the responding subsets of $\mathbb{^*N}$ should be internal. 
The idea behind this approach is that $h_1 << h_2$, so that $\displaystyle\sum_{k=0}^{h_1}$ can take a similar role to the sum  $\displaystyle\sum_{k\in\mathbb{N}}$. 
Therefore, we choose $h_1$ so that $\frac{h_1}{h_2} \approx 0$. We can do this without loss of generality, as we can find such an infinite hyperreal $h_1$ for any infinite hyperreal $h_2$ (e.g. $h_1 := \lfloor\sqrt{h_2}\rfloor$).
Now the argumentation translates rather analogue:
$$
\sum_{k=0}^{h_2} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!} = \bigg(\sum_{k=0}^{h_1} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!}\bigg) + 
\bigg(\sum_{k=h_1}^{h_2} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!}\bigg)
$$
$$
\sum_{k=0}^{{h_1}} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!} =
\sum_{k=0}^{{h_1}} \frac{({h_2}\cdot ({h_2}-1)\cdots({h_2}-k+1))}{{h_2}^k} \frac {x^k} {k!}
=\\
\sum_{k=0}^{h_1} \frac{{h_2}^k}{{h_2}^k} \frac {x^k} {k!} +\Delta_k= 
\sum_{k\in\mathbb{N}} \frac {x^k} {k!} + \Delta_k$$
Here, $\Delta_k$ is supposed to be the difference that is defined so that the equation holds.  We now have to show $\sum_{k=0}^{h_1} \Delta_k \approx 0$:
$$|\Delta_k| \underbrace{\le}_{\#1} \frac 2 {h_2} \cdot \frac {x^k}{k!} \underbrace{\le}_{\#2} \frac 2 {h_2} \implies \sum_{k=0}^{h_1} |\Delta_k| \le\frac{(h_1+1)\cdot 2} {h_2} \approx 0$$ 
#1: This one is a long calculation with a few estimations, but it should be rather clear. 
#2: This inequality holds for almost all $k$, as $\frac{x^k}{k!}$ tends to 0.
And therefore, $\sum_{k=0}^{h_1} \Delta_k \approx 0$ holds.
Finally, for the sum $\displaystyle \sum_{k=h_1}^{h_2} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!}$ we again use analogous arguments:
$$
\sum_{k=h_1}^{h_2} \frac{{h_2}!}{{h_2}^k({h_2}-k)!} \frac {x^k} {k!} \approx
\sum_{k=h_1}^{h_2} \frac{d_1 \cdot {h_2}!}{{h_2}^k({h_2}-k)!} \approx
\sum_{k=h_1}^{h_2} \frac{d_1 d_2}{({h_2}-k)!} =\\
d_1 d_2 \sum_{k=h_1}^{h_2} \frac{1}{({h_2}-k)!} =
d_1 d_2 \sum_{k=0}^{h_2-h_1} \frac{1}{k!} \approx d_1 d_2 e \approx 0
$$
 A: Your step (1) is not quite correct for several reasons:
A. While it is true that $\frac{(h\cdot (h-1)\cdots(h-k+1))}{h^k}\approx 1$, we have infinitely many summands in the sum you are working with, and so the relation of infinite proximity may not hold.
B. A more fundamental reason is that the expression $\sum_{k\in\mathbb{N}}$ is ill-defined.  Namely you are attempting to evaluate an infinite sum but the sum in question is not internal.  Therefore there is no way of defining it, since transfer is limited to internal statements.
Euler, of course, used arguments with infinite sums that can be justified with a bit of effort; see this 2017 publication in Journal of General Philosophy of Science.
The way to make it work is to use a hyperfinite sum with index stopping at $H\in {}^\ast\mathbb N \setminus \mathbb N$ (rather than sums over external index sets).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\lim_{n \to \infty}\sum_{k = 0}^{n}{n \choose k}\pars{x \over n}^{k} =
\sum_{k = 0}^{\infty}{x^{k} \over k!}:\ {\Large ?}}$.

Any way
$\ds{\bbx{\sum_{k = 0}^{n}{n \choose k}\pars{x \over n}^{k} =
\pars{1 + {x \over n}}^{n}}}$. However, the long way is:
\begin{align}
\sum_{k = 0}^{n}{n \choose k}\pars{x \over n}^{k} & =
\sum_{k = 0}^{\infty}{n \choose n - k}\pars{x \over n}^{n - k} =
\sum_{k = 0}^{\infty}{n \choose k}
\braces{\pars{n - k}!\bracks{z^{n - k}}\expo{xz/n}}
\\[2mm] & =
n!\bracks{z^{n}}\expo{xz/n}\ \overbrace{\sum_{k = 0}^{\infty}{z^{k} \over k!}}
^{\ds{\expo{z}}}\ =\
n!\bracks{z^{n}}\expo{\pars{1 + x/n}z}
\\[5mm] & =
n!\,{\pars{1 + x/n}^{n} \over n!} =
\pars{1 + {x \over n}}^{n}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\expo{x} =
\bbx{\sum_{k = 0}^{\infty}{x^{k} \over k!}}
\end{align}
