method of evaluating limit looks correct to me but solution obtained is wrong? why? \begin{equation}
 \lim_{n\to\infty} \left\lgroup n!/n^{n}\right\rgroup^{1/n}
\end{equation}
why is this solution wrong?
\begin{equation}
 n! = n(n-1)(n-2)......1\\
 \left\lgroup n!/n^{n}\right\rgroup=(n/n)\cdot(n-1)/n\cdot(n-2)/n.......1/n\\
\implies \left\lgroup n!/n^{n}\right\rgroup= 1\cdot(1-1/n)\cdot(1-2/n).......1/n\\
\implies \lim_{n\to\infty} \left\lgroup n!/n^{n}\right\rgroup^{1/n}= (1\cdot1\cdot1......\cdot1(n\ times))^{1/\infty}=1^0=1\\
\end{equation}
I know its wrong answer but don't know why.
thank you! :)
 A: As pointed out in the comments, you don't have $1^0$ but $0^0$ instead, because the last term in the product is $\frac{1}{n}$ and the other terms are less than $1$. So it is an indeterminate form that needes to be worked out wisely. In this case, by taking the $\log$.

But I don't see how to solve this using just elementary single-variable Calculus...
Let us use the weakest form of Stirling's approximation:
$$
\log n!
=
n \log n - n + e_n
,
$$
where
$$
0
\leqslant
e_n
\leqslant
2 + \log n
.
$$
(This form of Stirling's approximation can be proved using elementary Calculus, though.)
So taking $\log$ you get
$$
\log \lim_{n\to\infty} \Big(\frac{n!}{n^n}\Big)^{\frac{1}{n}}
=
\lim_{n\to\infty} \frac{\log n! - n \log n}{n}
=
\lim_{n\to\infty} \frac{-n + e_n}{n}
=
-1 + \lim_{n\to\infty} \frac{e_n}{n}
\overset{H}{=}
-1
,
$$
where in $\overset{H}{=}$ we used both l'Hôpital and Squeeze Theorem.
Finally,
$$
\lim_{n\to\infty} \Big(\frac{n!}{n^n}\Big)^{\frac{1}{n}} = \exp\bigg[ \log \lim_{n\to\infty} \Big(\frac{n!}{n^n}\Big)^{\frac{1}{n}} \bigg] = e^{-1}.
$$

Proof of Stirling's approximation. First write
$$\log n! = \log 1 + \log 2 + \cdots + \log n,$$
which
is bounded from below by
$$
\log n!
\geqslant
\int_0^n \log x \, d x
=
n \log n - n
$$
and from above by
$$
\log n!
\leqslant
\int_1^{n+1} \log x \, dx
=
(n+1) \log (n+1) - n
\leqslant
n \log n - n + \log n + (n+1)\log \frac{n+1}{n}
$$
A: Your approach has issues at many levels. Note that the last factor in the expression $n! /n^{n} $ is $1/n$, the second last factor is $2/n$ and third last factor is $3/n$ so when we take limits as you have done in your question we should get $0$'s in the end and according to your reasoning there should be $1$'s in the start. So ideally your approach should generate $$(1\cdot 1\cdots 0\cdot 0)^{1/\infty}$$ How do you know where one switches from $1$ to $0$ and further if there is a $0$ in product shouldn't the whole product be $0$? So this approach has some inconsistency.
Next problem is the deep and almost correct belief that limit of a complicated expression can be obtained from the limits of its sub-expressions in exactly the same manner as the complicated expression is obtained from its sub-expressions. This belief is more formally encoded in laws of algebra of limits and works if the sub-expressions are combined using a finite number of $+, -, \times, /$ symbols (some further conditions apply). In the current question the number of factors is $n$ which can't be considered finite as $n\to\infty$. Another aspect is that algebra of limits applies to arithmetical operations and with some effort can be extended to other operations (mainly those operations which are continuous). Thus the exponent in current question has limit $0$ but we are not sure of the limit of base and we don't know if the laws of limit apply to exponentiation in such a case.

As far as the solution of the current problem is concerned, the following theorem is a great help here:

Theorem: If $a_{n} $ is a sequence of positive terms and $\lim_{n\to\infty} a_{n+1}/a_{n}=L$ then $\lim_{n\to\infty} a_{n} ^{1/n}=L$.

For the current problem take $a_{n} =n! /n^{n} $ so that $a_{n+1}/a_{n}=(1+(1/n))^{-n}\to 1/e$. Hence the desired limit is $1/e$.
A: There are two ways to look at what is going on. In the question, the limit is viewed as
$$
\left(\prod_{k=1}^n\frac kn\right)^{1/n}
$$
where the product on the inside of the parentheses tends to $0$, not $1$ as computed in the question, and the exponent tends to $0$. This leads to the indeterminate form $0^0$. This means we need to be more careful; this approach doesn't work.
Another approach is to bring the exponent inside the product:
$$
\prod_{k=1}^n\left(\frac kn\right)^{1/n}
$$
where we are taking an infinite product of terms that tend to $1$. This is commonly known as the indeterminate form $1^\infty$. So we still have to apply more care.

To see that the limit is not $1$, as derived in the question, we have the bound
$$
\begin{align}
\prod_{k=1}^n\left(\frac{k}{n}\right)^{1/n}
&\le\prod_{k=1}^{\lfloor n/2\rfloor}\left(\frac12\right)^{1/n}\prod_{k=\lfloor n/2\rfloor+1}^n1^{1/n}\\
&=\left(\frac12\right)^{\frac{\lfloor n/2\rfloor}n}
\end{align}
$$
So we know the limit can be no greater than $\frac1{\sqrt2}$.

To compute the log of the limit, we can use a Riemann Sum:
$$
\begin{align}
\log\left(\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}\right)
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac{k}{n}\right)\\
&=\int_0^1\log(x)\,\mathrm{d}x\\[9pt]
&=-1
\end{align}
$$
In case there are concerns regarding the fact that $\log$ is not bounded on $[0,1]$, since $\log$ is monotonic, we have the bounds
$$
\int_0^1\log(x)\,\mathrm{d}x\le\frac1n\sum_{k=1}^n\log\left(\frac{k}{n}\right)\le\int_{1/n}^{1+1/n}\log(x)\,\mathrm{d}x
$$
and the Squeeze Theorem gives the limit claimed above.
