limits - How to evaluate $lim_{b to infty} left[frac{b^{t-1}}{e^b}right]$?

I encountered this limit when I studied the the gamma function. This limit arises when integrating the gamma function by parts. The author evaluates it as follows:

1. $\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty} \left\{\dfrac{\text{exp}[(t-1) \ln b]}{\text{exp}(b)}\right\}$


2. $\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty}\{\text{exp}[(t-1) \ln b-b]\}$


3. $\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty} \left\{\text{exp}\left[(t-1)b\left(\dfrac{ \ln b}{b}-1\right)\right]\right\}$

And finally, by l’Hôpital's rule: $$\dfrac{ \ln b}{b} = 0$$ then $$\lim\limits_{b \to \infty} e^{(t-1)b(0-1)} = \lim\limits_{b \to \infty} e^{-\infty}=0.$$

I can not make sense of the manipulation between step 2 and 3. Can someone offer an explanation? Thanks in advance.

Source: https://onlinecourses.science.psu.edu/stat414/node/142

Answer

$$\frac{e^{(t-1)\log b}}{e^b}=e^{(t-1)\log b-b}$$

and now just factor out $\;b\;$ in the exponent:

$$(t-1)\log b-b=b\left((t-1)\frac{\log b}b-1\right)$$

...and there's a mistake, either in your writing or in the book, as the factor $\;(t-1)\;$ does not multiply the whole thing, only $\;\log b\;$ ...