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Since $$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n={e}$$
My strong hunch is that the following statement must also be true $$\lim_{n\rightarrow \infty}\left(1+\frac{r}{n}\right)^n = {e^{r}}$$ for all $r>0$.
But I can neither prove or disprove it, any idea on how to prove it? Or if the statement is not true, how it should be modified so that it is true?
Your hunch is correct. Letting $u = \frac{n}{r}$, we have: $$\begin{align*} \lim_{n\to\infty}\left(1 + \frac{r}{n}\right)^n &= \lim_{n\to\infty}\left(\left(1+\frac{r}{n}\right)^{n/r}\right)^r\\ &= \lim_{u\to\infty}\left(\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= \left(\lim_{u\to\infty}\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= e^r. \end{align*}$$
Another way to see this. Suppose $$\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n = L.$$ Let us calculate $\ln(L)$:
$$\begin{align*} \ln(L) &= \ln\left(\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n \right)\\ &=\lim_{n\to \infty} \ln\left(\left(1+\frac{r}{n}\right)^n\right)\\ &=\lim_{n\to \infty} n\ln\left(1+\frac{r}{n} \right)\\ &=\lim_{n\to \infty} \frac{\ln\left(1+\frac{r}{n} \right)}{\frac{1}{n}}\\ &=\lim_{n\to\infty} \frac{\frac{1}{1+\frac{r}{n}}\cdot\frac{-r}{n^2}}{-\frac{1}{n^2}}\\ &=\lim_{n\to\infty} \frac{r}{1+\frac{r}{n}}\\ &=r, \end{align*}$$ where we have used the fact that $\ln(x)$ is continuous in $(0,\infty)$, and l'Hôpital's rule. Thus, $\ln(L)=r$, or equivalently, $L=e^r$.
