Euler's proof that $\lim_{n\to\infty} (1 + \frac x n)^n = \sum_{k=0}^n \frac{x^k}{k!}$, but over matrices and complex numbers Euler "proved" that the Taylor series and the binomial expression for $e^x$ are equivalent. The binomial expression for $e^x$ is $\lim_{n\to\infty} (1 + \frac x n)^n$, and the Taylor series is $\sum_{k=0}^n \frac{x^k}{k!}$. His "proof" uses the binomial theorem to expand the binomial expression, and then observes that the terms $n^{-k}{n \choose k}x^k$ approach $\frac{1}{k!}x^k$ for all $k$. He then concludes.
The Monotone Convergence Theorem for sequences of series can be used to justify this argument in the event that $x$ is non-negative.
Wikipedia shows how to generalise this to negative values of $x$ using the ad-hoc observation that $\left(1 - \frac r n \right)^n  \left(1+\frac{r}{n}\right)^n = \left(1-\frac{r^2}{n^2}\right)^n$.
What about complex values of $x$? Would it reduce to another ad-hoc observation that both functions satisfy $f(x+y)=f(x)f(y)$? What about matrix values of $x$?
I'd like an argument that still broadly resembles Euler's proof: The term $n^{-k}{n \choose k}x^k$ approaches $\frac{1}{k!}x^k$ as $n$ approaches $\infty$.
 A: We'll use the triangle inequality.
Pick any element $z$ of any unital Banach algebra $A$. $z$ could be a matrix or a complex number, for instance.
Let $\epsilon > 0$.
We can pick a value of $s$ such that the suffix of the Taylor series has magnitude at most $\epsilon/3$. In other words, we have that $\sum_{k=s+1}^\infty \frac{\vert z \vert^k}{k!} \leq \epsilon/3$. This follows from the fact that the Taylor series is convergent in the real case.
We then have that for any exponent $n$ in the binomial: $$\left|\left(1+\frac z n\right)^n - \sum_{k=0}^\infty \frac{z ^k}{k!}\right| \leq \sum_{k=0}^s \left|n^{-k}{n \choose k}z^k - \frac{z^k}{k!}\right| + \sum_{k=s+1}^\infty n^{-k}{n \choose k}\left|z\right|^k + \sum_{k=s+1}^\infty\frac{\left|z\right|^k}{k!}.$$
This is using the triangle inequality and the sub-multiplicative property of algebra norms. (Over complex numbers, the norm, written $|z|$, is multiplicative: $|wz|=|w||z|$, therefore also sub-multiplicative).
We break the right-hand side into its three summands: $$\begin{align}\sum_{k=0}^s \left|n^{-k}{n \choose k}z^k - \frac{z^k}{k!}\right| \tag{1},\\ \sum_{k=s+1}^\infty n^{-k}{n \choose k}\left|z\right|^k \tag{2}, \\\sum_{k=s+1}^\infty\frac{\left|z\right|^k}{k!} \tag{3}\end{align}.$$
For (3), we see that $$ \sum_{k=s+1}^\infty\frac{\left|z\right|^k}{k!} \leq \epsilon/3,$$ by fiat.
For (2), we see that $$\begin{align}
&n^{-k}{n \choose k}\left|z\right|^k  &= \frac1{k!}(1-\frac1n)(1-\frac2n)\cdots(1-\frac {k-1} n) \left|z\right|^k &\leq \frac{\left|z\right|^k}{k!},\\
\implies &\sum_{k=n+1}^\infty n^{-k}{n \choose k}\left|z\right|^k&&\leq\sum_{k=n+1}^\infty\frac{\left|z\right|^k}{k!}\\
&&&\leq \epsilon/3.\end{align}$$
(1) is eventually less than $\epsilon/3$ for large enough $n$ because $\frac{\left|z\right|^k}{k!} = \frac1{k!}(1-\frac1n)(1-\frac2n)\cdots(1-\frac {k-1} n) \to \frac{\left|z\right|^k}{k!}$.
Overall then, for large enough $n$, we see that the difference is at most $\epsilon$.
