Show that $\left(1+\dfrac{x}{n}\right)^n \to e^x$ as $n \to \infty$ in a normed ring $R$

Let's assume (as Landscape does) that we are in a Banach algebra. By the binomial theorem, one may write \begin{equation} \left(1+\frac{x}n\right)^n=\sum_{k=0}^\infty a_{k,n} \, ,\tag{$*$}\end{equation} where the $a_{k,n}$ are given by $$a_{k,n}=\left\{ \begin{matrix} \left(n\atop k\right) \frac{x^k}{n^k}&{\rm if}\; k\leq n\\ 0&{\rm if}\; k>n \end{matrix}\right.$$ For each fixed $k$, we have $\lim_{n\to\infty} a_{k,n}=\frac{x^k}{k!}$, because $\left(n\atop k\right)=\frac{n(n-1)\cdots (n-k+1) }{k!}\sim\frac{n^k}{k!}$. Moreover, we also have $\left(n\atop k\right)\leq \frac{n^k}{k!}$, so that $\Vert a_{k,n}\Vert\leq \frac{\Vert x\Vert^k}{k!}\cdot$ Since the series $\sum\frac{\Vert x\Vert^k}{k!}$ is convergent, we may apply the "dominated convergence theorem for series" to let $n\to\infty$ in $(*)$. This gives $$\lim_{n\to\infty} \left(1+\frac{x}n\right)^n=\sum_{k=0}^\infty \frac{x^k}{k!}=e^x\, .$$