Is the denominator really $2n!$ and not $(2n)!$?

Tintarn wrote: Is the denominator really $2n!$ and not $(2n)!$? You are right.

By Power Series/Taylor Series, it is clear that \[\frac{f^{(2n)}{1}}{e \cdot (2n)!}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\dots -\frac{1}{(2n-1)!}+\frac{1}{(2n)!}\]so the limit in question is just \[e\cdot \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}=e \cdot \frac{1}{e}=1\]

Another way to prove it would be to consider the function $f(x)-\frac{1}{x}=\frac{e^x-1}{x}$. This is holomorphic at $x=0$ and hence entire, so the power series at any point, in particular at $x=1$ must converge, so the coefficients of that power series must go to zero. Hence the limit of the coefficients of the series for $f$ is just the same as for the coefficients of the series of $\frac{1}{x}$ around $x=1$. But that's just the geometric series $\frac{1}{1+y}=1-y+y^2-y^3+\dots$ so the limit is $1$ (since we are only considering the even exponents).

Or just good old-fashioned induction. Note that $xf(x)=e^x$. Observe the pattern $$xf^{(1)}(x) + f^{(0)}(x)=e^x$$$$xf^{(2)}(x) + 2f^{(1)}(x)=e^x$$$$xf^{(3)}(x) + 3f^{(2)}(x)=e^x$$$$\vdots$$$$xf^{(n)}(x) + nf^{(n-1)}(x)=e^x$$Of course, it is easy to prove via induction. This implies, on setting $x=1$, that for all naturals $n$, we have $$f^{(n)}(1) + nf^{(n-1)}(1)=e$$To make a nice telescoping, rewrite this as $$\frac{(-1)^n f^{(n)}(1)}{n!} - \frac{(-1)^{n-1} f^{(n-1)}(1)}{(n-1)!} = \frac{(-1)^n \cdot e}{n!}$$Adding this from $n=1$ to $n=2m$, we get $$ \frac{f^{(2m)}(1)}{(2m)!} = e \cdot \sum_{i=0}^{2m} \frac{(-1)^i}{i!}$$so that $$\lim_{m \to \infty} \frac{f^{(2m)}(1)}{(2m)!} = 1$$

Note that \begin{align*} \frac{e^{x+1}}{1+x} & =e\cdot \left (\sum_{n\geq 0} \frac{x^n}{n!}\right)\left(\sum_{n\geq 0}(-1)^n x^n\right) \\ &=e \sum_{n\geq 0} \left (\sum_{0\leq k \leq n} \frac{(-1)^{n-k}}{k!}\right) x^n.\end{align*}So, $\frac{e^x}{x}= \sum_{n\geq 0} e\left (\sum_{0\leq k \leq n} \frac{(-1)^{n-k}}{k!}\right) (x-1)^n$. So by Taylor's Theorem, $$\frac{f^{2n}(1)}{(2n)!}=e \sum_{0\leq k \leq n} \frac{(-1)^{2n-k}}{k!}=e \sum_{0\leq k \leq n} \frac{(-1)^{k}}{k!}\longrightarrow e \sum_{k\geq 0} \frac{(-1)^{k}}{k!}=e \cdot \frac 1e=1.$$