Came across a book that assumes we understand why this holds:
$\lim \limits_{n \to \infty}(1+\frac rn)^n = \lim \limits_{n \to \infty}(1+\frac1n)^{rn}$
I already verified both sides converge numerically but I'd like to know if there's a more rigorous proof. This probably is a duplicate but I had a hard time figuring out what to look for.
We see that \begin{align} \lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n&=\lim_{rm\to\infty}\left(1+\frac{r}{rm}\right)^{rm}\\ &=\lim_{rm\to\infty}\left(1+\frac{1}{m}\right)^{rm}\\ &=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{rm}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{rn}\\ \end{align} note that the steps from the first line and the second to the third line can only be done if $r>0$; I leave the case $r<0$ and $r=0$ to you (spoilers: case $r=0$ is simply filling in $r=0$ and you're done, and $r<0$ is painfully similar to the $r>0$ case).
There's actually a general rule:
If $f(0)=1$ and $f'(0)=0$ then $$\lim_{n\to\infty} f\left(\frac 1n\right)^n=1.$$
Proof: First, note that $n\left(f\left(\frac{1}{n}\right)-1\right)\to f'(0)=0$. Then, using: $$ f\left(\frac{1}{n}\right)^n=\left(1+\left(f\left(\frac{1}{n}\right)-1\right)\right)^n=1+\sum_{k=1}^{n}\binom{n}{k}\left(f\left(\frac{1}{n}\right)-1\right)^k$$ and that $\binom{n}{k}\leq n^k$, you have:
$$\left|f\left(\frac{1}{n}\right)^n-1\right|<\sum_{k=1}^{n}\left|n\left(f\left(\frac{1}{n}\right)-1\right)\right|^k\tag{1}$$
When $|n(f(1/n)-1)|<\epsilon<1$ then the right side is bounded above by:
$$\sum_{k=1}^{\infty} \epsilon^k =\frac{\epsilon}{1-\epsilon}$$
Since $n(f(1/n)-1)\to 0$, this means the right side or (1) converges to zero, and thus $f(1/n)^n\to 1.$
Then let $f(x)=\frac{(1+rx)}{(1+x)^r}$. Then $f(0)=1$ and $f'(x)=\frac{r(1+x)-r(1+rx)}{(1+x)^{r+1}}=\frac{(r-r^2)x}{(1+x)^{r+1}}$ and hence $f'(0)=0$.
