Determine $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$ and $\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$ I want to use Stirlings Formula $\lim_{n\to \infty} {n!} \sim \lim_{n \to \infty} \frac{n^n}{e^n}\cdot \sqrt{n}$ to evaluate the following limits:
$$\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$$
$$\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$$
For the first one I feel like some algebraic manipulation would yield $\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} = \frac{e}{n^{^1/n}}$ which would imply $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}} = e$ but can I just do equivalence transformation with an asymptotical equality, as opposed to a precise equality?
The second one looks like Taylor Series development, but I don't see the path...

If you are unsure about what you are allowed to do with equivalents, then go back to the equivalent characterization as first-order Taylor expansion (in particular, when you write Stirling's formula, you should not have $\lim$ on either side: this is mixing two different notions; you are also missing a constant):
$$
n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\frac{n^n}{e^n} \quad\Longleftrightarrow\quad n! = \sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right) \tag{1}
$$
With (1), we can then write
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{n}{\sqrt[n]{\sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right)}}= \frac{n}{\frac{n}{e}(2\pi n)^{1/2n}\sqrt[n]{1 + o\left(1\right)}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
$$
Now, we can observe that
$$
(2\pi n)^{1/2n} = \exp\left(\frac{1}{2n}\ln(2\pi n)\right) \xrightarrow[n\to\infty]{} e^0 = 1
$$
and
$$
(1+o(1))^{1/n} = \exp\left(\frac{1}{n} \ln(1+o(1))\right) \xrightarrow[n\to\infty]{} e^0 = 1$$
so that
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
\xrightarrow[n\to\infty]{} \boxed{e}
$$

For the second: establishing that $$\sum_{k=1}^{n-1} \log k = \log \prod_{k=1}^{n-1} k = \log((n-1)!)$$ and applying Stirling's formula will do the trick:
$$
\log n!  = n\log n - n + \frac{1}{2}\log (2\pi n) + o(1) \tag{2}
$$
from which
$$\begin{align}
\frac{n+\log( (n-1)! )}{\log n} +1 - n
&= 
\frac{n+\log( n! ) - \log n}{\log n} +1 - n
= 
\frac{n+\log( n! )}{\log n} - n
\\
&= \frac{n\log n + \frac{1}{2}\log (2\pi n) + o(1)}{\log n} - n\\
&= \frac{\frac{1}{2}\log (2\pi n) + o(1)}{\log n} \xrightarrow[n\to\infty]{} \cdots
\end{align}$$
