Does $n!$ divide $ n^n$?

Question

Today while I was reading on how to shuffle an array I came across a statement that claims we shall not swap an array entry with the whole array range when shuffling the array otherwise we end up with a biased result. Click here for details.

They demonstrated that by saying that $n!$ does not divide $n^n$ because $(n-1)$ divide $n!$ but at the same time $(n-1)$ does not share any prime factor with $n$, if I understood it right.

However, I did not grasp the "proof" given, so, can you help me figure out how to demonstrate it in a more rigorous approach?

Answer 1

In the comments it is detailed how to find a prime $p$ such that $p \; \mid \; n-1$ but $p \; \nmid \; n^n$.

Suppose it were true now that $(n-1)! \; \mid \; n^n$. Then we would have $$ p \mid n -1 \mid (n-1)! \mid n^n $$ so $p \mid n^n$, which is impossible.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With Induction, you will use: $$ {\pars{n + 1}^{n + 1} \over \pars{n + 1}!} = {\pars{n + 1}^{n} \over n!} ={n^{n} \over n!}\,{\pars{n + 1}^{n} \over n^{n}} ={n^{n} \over n!}\ \underbrace{\pars{1 + {1 \over n}}^{n}} _{\ds{\not\in\ {\mathbb N}_{+}}} $$ $\imp\quad n!\ \not\vert\ n^{n}$

Answer 3

No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 2^3 \cdot 3$.