product of numbers raised to binomial coefficients I want to find a closed form expression for $\prod_{i = 1}^{n} i^{\binom{n}{i}}$. If a closed form expression is not available, can a recursive relation be formed? Please help.
Regards,
Akhil.
 A: This is  OEIS sequence A229333. 
More generally, if $$A(n,m) = \prod_{i=m}^{n} i^{n\choose {i-m}}$$
so that your expression is $A(n,0)$, we have $$A(n+1,m) = (n+1)^{{n+1}\choose m} A(n,m) A(n,m+1)$$
with $A(n,n) = n$
A: Take a logarithm from both side and we get:
$$
\sum_{i = 1}^{n} {\binom{n}{i}}\log i=\sum_{i = 1}^{n} {\binom{n}{i}}\log\frac{i}{n}+\log n\sum_{i = 1}^{n} {\binom{n}{i}}=\sum_{i = 1}^{n} {\binom{n}{i}}\log\frac{i}{n}+2^n\log n.
$$
Now consider Bernoulli random variables $X_1,...,X_n$ such that $P(X_i=1)=x$. Define $S_n=X_1+...+X_n$. Then we can write down the following:
$$
\mathbb E\left(\log(\frac{S_n}{n})\right)=\sum_{i = 1}^{n} {\binom{n}{i}}x^i(1-x)^{n-i}\log\frac{i}{n}
$$
By the law of large numbers, $\mathbb E\left(\log(\frac{S_n}{n})\right)$ will be close to $\log(x)$ as $n\to\infty$. $\mathbb E\left(\log(\frac{S_n}{n})\right)$ is a polynomial called Bernstein polynomial, denoted by $B_n(x)$. Bernstein's theorem says that $B_n(x)$, corresponding to the continuous function $f(x)$ converges uniformly to $f(x)$.
Now consider the following:
$$
B_n(\frac{1}{2})=\sum_{i = 1}^{n} {\binom{n}{i}}2^{-n}\log\frac{i}{n}
$$
where $B_n(x)$ is Bernstein polynomial for $f(x)=\log(x)$. Now using Brenstein theorem we can say that $B_n(\frac{1}{2})$ converges uniformly to $\log(\frac{1}{2})$. Hence we have:
$$
\prod_{i = 1}^{n} i^{\binom{n}{i}}=\exp\left(2^n\log n+2^n B_n(\frac{1}{2})\right)\approx \exp\left(2^n\log n+2^n \log(\frac{1}{2})\right)
$$
Finally we can approximate the previous one by:

$$
\prod_{i = 1}^{n} i^{\binom{n}{i}}\approx \left( \frac{n}{2}\right)^{2^n}
$$

