in limits for 1^∞ type of indeterminate form, the direct solution is e^{g(x)[f(x)-1]}
I am not sure what you mean by "direct solution" but you can convert the form so L'hosiptials rule can be used.
If $\displaystyle \lim_{x \to \infty} f(x)=1$ and $\displaystyle \lim_{x \to \infty} g(x)=\infty$ then
$\displaystyle [f(x)]^{g(x)} = \exp [g(x) \ln(f(x))]= \exp \left[ \dfrac{\ln(f(x))}{\frac{1}{g(x)}}\right]$
This is now in the form $\displaystyle \dfrac{0}{0}$ so you can apply L.H.
Note the fraction could also be arrainged to get infinity over infinity