1 to the power of infinity formula

There is a general formula for indeterminate form $1 ^ {\infty}$ which I'm looking for a proof which is also used here. (picture)

Given $$\lim_{x\to a} f(x) = 1$$ and $$\lim_{x\to a} g(x) = \infty$$, what is $$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}}\quad ? $$

I would appreciate it if somone could give me a proof of this formula.

I believe this is where the identity is coming from.

\begin{align} \lim_{x\to a}f^g &= \lim_{x\to a}(1+f-1)^g \\ &= \lim_{x\to a}\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^g \\ \\ &= \lim_{x\to a}\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{g\frac{f-1}{f-1}} \\ \\ &= \lim_{x\to a}\left[\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{\frac{1}{f-1}}\right]^{g(f-1)} \\ \\ &= \lim_{x\to a}\left[\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{\frac{1}{f-1}}\right]^{\lim_{x\to a}g(f-1)} \qquad (*) \\ \\ &= e^{\lim_{x\to a}g(f-1)} \end{align}

Where the first limit is a form of the limit definition of $e$. I put a note (*) next to one step I am uneasy about. I am unsure why we can separately evaluate limits here. Perhaps someone else can comment on this.

Remember that $f$ and $g$ are functions of $x$, so to be more precise, we should write $f(x)$ and $g(x)$ instead of $f$ and $g$. This applies to the answer below and to the other answers which have also adopted the shorthand used in the question of $f$ for $f(x)$ and $g$ for $g(x)$.

If $\lim\limits_{x\to a}f=1$, then $$ \begin{align} \log\left(\lim_{x\to a}f^g\right) &=\lim_{x\to a}\log\left(f^g\right)\\[6pt] &=\lim_{x\to a}\log(f)\,g\\ &=\lim_{x\to a}\frac{\log(f)}{f-1}\lim_{x\to a}\,(f-1)\,g\\[6pt] &=\lim_{x\to a}\,(f-1)\,g \end{align} $$ Therefore, $$ \lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right) $$