I found one definition of $e$ to be $\lim_{n\to \infty}(1 + \frac{1}{n})^n = e$ and I checked it with a graphing calculator to be true. However the $1$ in the numerator is a special case, where it can be replaced with a different number and it affects the power to which $e$ is raised.
I could not find anyone prove this specifically, I found one proving the $b = 1$ case (1233dfv's answer), which I thought I could use to prove the $b = b$ case.
My go at it:
$y = \lim_{n\to \infty}(1 + \frac{b}{n})^n$
$\ln(y) = \lim_{n\to \infty}n\ln(1 + \frac{b}{n})$ (indeterminate form "$\infty\times 0$")
$\ln(y) = \lim_{n\to \infty} \ln(1+\frac{b}{n})/(\frac{1}{n})$ (indeterminate form "$0 / 0$", can use L'Hopital rule)
$\ln(y) = \lim_{n\to \infty}\frac{\frac{1}{1+\frac{b}{n}}\cdot\frac{-b}{n^2}}{\frac{-1}{n^2}}$
$\ln(y) = \lim_{n\to \infty} \frac{1}{1+\frac{b}{n}} \cdot b$
Since $b$ is a number and infinity is not, $b/n = 0$, thus
$\ln(y) = b$
$\ln(y) = \ln(e^b)$
$y = e^b$
Are my calculations and reasoning correct?
