Proving that exponential growth at rate r equals exponential growth at rate 1 to the power of r This is a fairly basic result, but I could not find anything about it here.

How do you prove that the following relationship exists, and where does the basis for it originate from:

$$(1+r/∞)^∞=\left( \left( 1+\frac{1}{\infty}\right) ^{\infty}\right)^r$$



Improved by hjhjhj57:

$$\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^r$$



That is to say, a constantly compounded interest at the period's interest rate $r$ is equivalent to a constantly compounded interest at a period's rate of $100$ % to the power of $r$. 

The second formula is equivalent to the exponent function $e^r$. Can other functions with the same features (namely that the $f(dx) = f(x)$), be derived from the formula by changing the $r$ and maybe with some other alterations? I am asking this since using the $100$ % interest rate as a definition for $e$ seems somewhat arbitrary and I am wondering if other functions with the same features can be derived from the formula.

EDIT: The idea of this question is not necessarily to prove the equation above holds, but to explain where the relationship or the intuition comes from in the first place or how it was first discovered. In other words why does $e^r$ represent growth factor when the period's rate is $r$ and added constantly, given that $e^1$ equals period's rate of $100$%.

The second question was already answered here: Function is equal to its own derivative.
Consider the derivative of the differences with respect to r.

$\frac{1}{n}(1+\frac{r}{n})^{n-1}-nlog((1+\frac{1}{n})^{nr})(1+\frac{1}{n})^{nr-1}$

This expression tends towards 0 as n tends towards infinity so they are equal to each other.
Consider the derivative of the differences with respect to r.

$\frac{1}{n}(1+\frac{r}{n})^{n-1}-nlog((1+\frac{1}{n})^{nr})(1+\frac{1}{n})^{nr-1}$

This expression tends towards 0 as n tends towards infinity so they are equal to each other.