What are the steps in breaking down the exponent in this limit analysis? I'm trying to understand the reasoning in the following step of a limit analysis:
$$\lim_{n \to \infty} n\left(\left[1- \frac{1+c}{\frac{n}{\ln(n)}} \right]^{\frac{n}{\ln (n)}}\right)^{(n-1)\ln n/n} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}$$
I understand the "inner" part; $\lim_{n \to \infty} [1-\frac{1+c}{\frac{n}{\ln (n)}}]^{\frac{n}{\ln n}} = e^{-(1+c)}.$ And I sort of see that outer exponent $((n-1)\ln n) /n = (\ln n) - (\ln n / n)$ and the second part goes to 0, but it's not clear to me what rules actually justify "bringing the limit to the exponent". What are the actual steps involved in deducing this limit?
More generally, these types of asymptotic analysis show up in comp sci all the time and I feel there is a bag of tricks that I am missing.
 A: You can't in general replace a sub-expression by its limit while evaluating the limit of a complicated expression. There are ways to do so in very specific circumstances when the sub-expression is a term or a factor in the whole expression, but otherwise this is not allowed.
A better strategy is to take logarithm and thus if $L$ is the desired limit then we have by continuity of logarithm $$\log L=\lim_{n\to\infty} \log n+(n-1) \log\left(1-\frac{(1+c)\log n} {n} \right) $$ and the expression under limit above can be written as $$\log n+n\log\left (1-\frac{(1+c)\log n} {n} \right) - \log\left(1-\frac{(1+c)\log n} {n} \right) $$ Clearly the last term tends to $0$ and hence we get $$\log L=\lim_{n\to\infty} \log n+n\log\left(1-\frac{(1+c)\log n} {n} \right) \tag{1}$$ If $t=(\log n)/n$ then we can write $$\log L=\lim_{n\to \infty} \log n\cdot\frac{t+\log(1-(1+c)t)}{t}$$ The second factor above tends to $-c$ and the fraction can thus be replaced by its limit $-c$ as long as $c\neq 0$.
Thus if $c\neq 0$ we get $$\log L=-\lim_{n\to\infty} c\log n$$ which means that $L=0$ if $c>0$ and $L=\infty$ if $c<0$.
If $c=0$ then we have more work to do. In that case $$\log L=\lim_{n\to\infty} \log n\cdot t\cdot\frac{t+\log(1-t)}{t^2}$$ and last factor tends to $-1/2$ and then we have $$\log L=-\frac{1}{2}\lim_{n\to\infty }  \frac{(\log n) ^2}{n}=0$$ so that $L=1$ in this case.

Most cases of asymptotic analysis during evaluation of limit are nothing more than omission of valid steps (on the assumption that they are obvious to the reader) and may appear as tricks. But in reality they are nothing more than application of limit laws.
The specific treatment in your question is nothing more than applying Taylor series in equation $(1)$ and discarding all terms except the first because all the other terms tend to $0$. Thus we get from $(1)$ $$\log L=\lim_{n\to\infty } \log n-n\cdot(1+c)\frac{\log n} {n} $$ and thus $$L=\lim_{n\to\infty} ne^{-(1+c)\log n} =\lim_{n\to\infty} n^{-c} $$
A: For this limit, you can bring up the $\ln(n)$ from the denominator, and then make the exponent match by using $(a^b)^c = a^{bc}$:
$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{\frac{n}{\ln (n)}}\right)^{\frac{(n-1)\ln n}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)^{\frac{1}{\ln(n)} \cdot \frac{(n-1)\ln n}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)^{\frac{n-1}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}
$$
This particular problem can be simiplied further into a form that uses the common limit property $\lim_{x \to a}\left[\frac{f(x)}{g(x)}\right] = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}$, so that we don't have to worry about whether a corresponding rule holds for exponents.
$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n-1}\right) = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\frac{\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)}{\lim_{n \to \infty} \left[1- \frac{(1+c)\ln(n)}{n} \right]} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\frac{\lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}}{1-0}
$$

Breaking down that last step for the numerator requires the chain rule for limits, which states that if $\lim_{u \to b} f(u) \to L$ and $\lim_{x \to a} g(x) = b$, and $f(x)$ is continuous at $x=b$, then $\lim_{x \to a} f(g(x)) \to L$.  Colloquially, as the input to a function approaches some input limit, its output approaches some output limit, and we can link up the output limit of one function to be the input limit of another function.
$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right) \\
\lim_{n \to \infty} n \cdot \lim_{n \to \infty} \left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right) \\
$$
Extracting out the inner function, define:
$$
g(n) = (1+c)\ln n \\
\lim_{n \to \infty} g(n) \to G
$$
And now we can write our expression as:
$$
\lim_{n \to \infty} n \cdot \lim_{x \to G}\left[\lim_{n \to \infty} \left(\left[1- \frac{G}{n} \right]^{n}\right)\right] \\
$$
But whatever G may end up being, this portion of behavior as $n \to \infty$ is well defined as $e^x$, whatever the argument $x$ approaches.  G need not vary with n for this next step:
$$
\lim_{n \to \infty} n \cdot \lim_{x \to G} e^G
$$
Running the chain rule back the other direction, substituting back in for G:
$$
\lim_{n \to \infty} n \cdot \lim_{n \to \infty} e^{(1+c)\ln n}
$$
And running the product rule in reverse, we have the desired
$$
\lim_{n \to \infty} ne^{(1+c)\ln n}
$$
