Disguised as something?

$(n!)^{\frac{1}{n}} = (n(n-1)(n-2)...1)^{\frac{1}{n}} \le \left(n^n\right)^{\frac{1}{n}} = n$ $\implies \dfrac{n+1}{(n!)^{\frac{1}{n}}} \ge \dfrac{n+1}{n} \space$ and minimum happens when $(n!)^{\frac{1}{n}} = n \implies n = 1\space$ and $\dfrac{n+1}{(n!)^{\frac{1}{n}}} = \dfrac{1+1}{(1!)^1} = 2 \implies 2^2 = \boxed{4}$ .

Sum of first $n$ positive integers is $\dfrac{n(n+1)}{2}$ .

By AM-GM inequality ,

$\dfrac{1 + 2+ 3 \cdots + (n-1) + n}{n} \geq (1\cdot 2\cdot 3\cdot \cdots (n-1) \cdot n )^{\frac{1}{n}}$

$\implies \dfrac{n(n+1)}{2n} \geq (n!)^{\frac{1}{n}}$

$\implies \dfrac{(n+1)}{2} \geq (n!)^{\frac{1}{n}}$

$\implies \dfrac{(n+1)}{(n!)^{\frac{1}{n}}} \geq 2$

$\implies$ minimum value of $\dfrac{(n+1)}{(n!)^{\frac{1}{n}}}$ is $2$ , hence our answer is $\boxed{4}$ .

Let $x$ be the or representative of the given expression.

Then $x^{n} = n+1(n+1)^{n-1}/n!$

The value of $(n+1)^{n-1}/n!$ will increase as with respect to $n$ therefore to get minimum x we have to put minimum value for n which is 1.

Harsh Shrivastava I know that you can provide a better solution. ;)

We can't have negative factorials here, that's for sure.

After, negative integers, comes $0$ .We could keep $n$ as $0$ , but then the power of $\frac{1}{0}$ would not work.

So we keep $n$ as $\boxed {1}$ and $\large \displaystyle \text {Bingo !}$

I bet everyone who solved this till 4/4/15 used this method hence nobody has written a solution till now.