Non-calculus ways to get higher-order terms in $(1 + k/n)^n$? There's a well-known limit $$\lim_{n \to \infty} \left( 1 + \frac{k}{n} \right)^n = e^k $$ that can be proved either by computing limits of individual terms in the binomial expansion, or by defining $$f(x) = (1 + kx)^{1/x}$$ and using calculus to compute $$\lim_{x \to 0} \ln f(x) = \lim_{x \to 0} \frac{\ln (1+kx)}{x} = \lim_{x \to 0} \frac{k}{1+kx} = k.$$
I was working on a problem for which the rate of convergence of this limit is relevant: computing the expected size of the image of a random function from $\{1, \ldots, n\}$ to itself. The chance that any particular number is included in the image is $1 - \left(\frac{n-1}{n}\right)^n$, so the expected size, by linearity of expectation, is $n - n\left( \frac{n-1}{n} \right)^n = \left( 1 - \frac{1}{e} \right) n + O(1).$ WolframAlpha showed that the error term was a constant $\frac{2}{e} + O(n^{-1})$, and I proved this myself by developing $f$ as a power series:
\begin{align*}
\ln f(x) &= \frac{\ln (1+kx)}{x} \\
\frac{d}{dx} \ln f(x) &= -\frac{\ln (1+kx)}{x^2}+ \frac{k}{x (1+kx)} \\
f'(x) &= f(x) \frac{d}{dx} \ln f(x) \\
&= (1 + kx)^{1/x}  \left[-\frac{\ln (1+kx)}{x^2}  + \frac{k}{x (1+kx)}\right] \\
\lim_{x \to 0} f(x) &= e^k \lim_{x \to 0}  \left[-\frac{\ln (1+kx)}{x^2} + \frac{k}{x (1+kx)}\right]  \\
&= e^k \lim_{x \to 0} \frac{-(1+kx) \ln (1+kx) + kx}{x^2 (1+kx)} \\
&= e^k \lim_{x \to 0} \frac{-k \ln (1+kx) }{3kx^2 + 2x} \\
&= e^k \lim_{x \to 0} \frac{-k^2}{(6kx+2) (1+kx)} \\
&= \frac{-e^k k^2}{2}
\end{align*}
and therefore $$\left(1 + \frac{k}{n}\right)^n = e^k - \frac{e^k k^2}{2} \frac{1}{n} + O\left( \frac{1}{n^2} \right).$$
(This procedure can be continued, cumbersomely, to get higher-order terms, each of which is a $e^k$ times a polynomial in $k$ with rational coefficients that don't seem to match any OEIS sequence.) I'd like to know, though: Is there any way of getting the $-e^k k^2/2n$ term purely through series manipulation, without resorting to calculus? I tried finding a way for a bit, but the computations quickly get unpleasant.

Consider
$$a_n=\left(1 + \frac{k}{n}\right)^n \implies \log(a_n)=n \log\left(1 + \frac{k}{n}\right)$$ Using Taylor for large values of $n$
$$\log(a_n)=k-\frac{k^2}{2 n}+\frac{k^3}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor
$$a_n=e^{\log(a_n)}=e^k\left(1-\frac{k^2}{2 n}+\frac{k^3 (3 k+8)}{24
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$
