It's a shorthand notation, but is not really "bad form" if both you and your readers understand what is really going on. What you really mean at each step where you write something like "$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} g(n)$" is
"if $\lim_{n \to \infty} g(n)$ exists, then $\lim_{n \to \infty} f(n)$ exists and has the same value". When at the end of the calculation you find that the last limit does exist,
that tells you that everything is good and you have found the limit you wanted. If you
ended up with a limit that doesn't exist, then you might not know about the original limit.
EDIT:
In particular, I note that you are using l'Hopital's rule to go from $\lim_{n \to \infty} \log\left(\frac{n-1}{n}\right)/(1/n)$ to $\lim_{n \to \infty} (1/(n^2-n))/(1/n^2)$. This is OK since the rule says if $f(n)$ and $g(n)$ both go to $0$ or $\infty$ as $n \to \infty$ and $\lim_{n \to \infty} f'(n)/g'(n) = L$ exists, then $\lim_{n \to \infty} f(n)/g(n) = L$ as well.
You shouldn't use the rule in the other direction, since it can happen that $\lim_{n \to \infty} f(n)/g(n)$ exists but $\lim_{n \to \infty} f'(n)/g'(n)$ does not.
I am going to calculate your limit without using Taylor series but only the two following notable limits:
$\lim\limits_{x\to0}\dfrac{\ln(1+x)}{x}=1\;,\quad\lim\limits_{x\to0}\dfrac{x-\ln(1+x)}{x^2}=\dfrac12\;.$
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^2\ln\left(1+\dfrac1x\right)^x\right]=$
$=\lim\limits_{x\to\infty}\dfrac{\left(1+\frac1x\right)^x-e\ln\left(1+\frac1x\right)^x}{\left[1-\ln\left(1+\frac1x\right)^x\right]^2}\cdot\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\frac1x\right)^x\right]^2$
Now I am going to calculate the first limit by using the following substitution:
$t=\dfrac1e\left(1+\dfrac1x\right)^x-1\;.$
$\lim\limits_{x\to\infty}\dfrac{\left(1+\dfrac1x\right)^x-e\ln\left(1+\dfrac1x\right)^x}{\left[1-\ln\left(1+\dfrac1x\right)^x\right]^2}=$
$=\lim\limits_{t\to0}\dfrac{e(1+t)-e\ln\big[e(1+t)\big] }{\left\{1-\ln\big[e(1+t)\big]\right\}^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{1+t-1-\ln(1+t)}{\big[1-1-\ln(1+t)\big]^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{\ln^2(1+t)}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{t^2}\cdot\lim\limits_{t\to0}\left[\dfrac{t}{\ln(1+t)}\right]^2=$
$=e\cdot\dfrac12\cdot1^2=\dfrac e2\;.$
Now I am going to calculate the second limit by using the following substitution:
$u=\dfrac1x\;.$
$\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\dfrac1x\right)^x\right]^2=$
$=\lim\limits_{x\to\infty}\left[x-x^2\ln\left(1+\dfrac1x\right)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac1u-\dfrac1{u^2}\ln\big(1+u\big)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac{u-\ln(1+u)}{u^2}\right]^2=$
$=\left(\dfrac12\right)^2=\dfrac14\;.$
Consequently,
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\dfrac e2\cdot\dfrac14=\dfrac e8\;.$
Best Answer
\begin{align*} \lim\limits_{x \to \infty} f(x)^{g(x)} &= \lim\limits_{x \to \infty} \exp\left(\log(f(x)^{g(x)})\right)\\ &= \lim\limits_{x \to \infty} \exp\left(g(x)\log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x)\log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x) \cdot \lim\limits_{x \to \infty} \log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x) \cdot \log(\lim\limits_{x \to \infty} f(x))\right)\\ &= \ldots \\ &= \lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty} g(x)}, \end{align*} so what you need is the existence of the limits and $\lim\limits_{x \to \infty} f(x) > 0$.