# “Closed” form for $\prod\limits_{i=1}^{n} i^{[(i,n)=1]}$?

How could we derive a closed form for $\prod\limits_{i=1}^{n} i^{[(i,n)=1]}$?

Here "$[s]$" is the Iverson bracket and "$(a,b)$" is the greatest common divisor.

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I don't think it is derivable but I would call it an Eulerial. :) I guess some bounds can be derived from those that are known for the totient function itself. mathoverflow.net/questions/65390/… –  Dan Brumleve Oct 11 '11 at 9:14
More a restatement than a closed form, but: $\prod\limits_{k=1}^{n-1} k^{[\gcd(k,n)=1]}$ would be a compact way of putting it. Here $[p]$ is the Iversonian bracket. –  Ｊ. Ｍ. Oct 11 '11 at 9:28
BTW: there doesn't seem to be a closed form, but the product of all totatives of a number is listed in the OEIS. –  Ｊ. Ｍ. Oct 11 '11 at 9:37
@J.M:Wow,thanks for updating me with those names (totatives/iverson),Btw is it iversonian or iverson bracket ... –  Quixotic Oct 11 '11 at 9:41
"Iversonian bracket" and "Iverson bracket" are the same thing. :) –  Ｊ. Ｍ. Oct 11 '11 at 9:45

This is called a Gauss Factorial, and is used frequently in Modular arithmetic. It satisfies a theorem analogous to Wilson's Theorem, and can be also be thought of the $p$-adic gamma function.
Define $$N_n !:=\prod_{{1\leq j\leq N}\atop{\gcd(j,n)=1}} j.$$
We have the following congruence: $$(n-1)_n!\equiv \begin{cases} -1\pmod n & \text{for }n=2,4,p^{\alpha},p^{2\alpha}\\ 1\pmod n & \text{otherwise} \end{cases}$$
Lastly, $$\Gamma_p(z)=\lim_{N\to z} (-1)^N (N-1)_p !$$ where $N$ runs through any sequence of positive integers $p$-adically approaching $z$.
Remark: If you want some other form, inclusion exclusion implies that for squarefree $n$, $$n_n! =n!\prod_{d|n}\left(d^{\frac{n}{d}}\left(\frac{n}{d}\right)!\right)^{\mu(d)},$$ where $\mu$ is the Möbius function.