Identity with a product

Question

I have some difficulties to show the following identity

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{(nm/2)-m}$$

My attempt:

$$\prod_{i=1}^m 2^{\frac{n-i-1}{2}}\triangleq 2^k$$ where $$\begin{aligned} k&\triangleq\sum_{i=1}^m \frac{n-i-1}{2}=\frac{1}{2}\sum_{i=1}^n (n-i-1)\\ &=\frac{1}{2}\left[(n-1)\sum_{i=1}^m1-\sum_{i=1}^m i\right]\\ &=\frac{1}{2}\left[(n-1)m-\frac{m(m+1)}{2}\right]\\ &=\frac{nm}{2}-\frac{m^2+3m}{4} \end{aligned}$$

so my conclusion is,

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{nm/2-(m^2+3m)/4}$$

Where am I wrong?

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Update:

I have inferred the previous equality from the following equation $$ \prod_{i<j}^m [(2\pi)^{1/2}]\prod_{i=1}^m\left\{2^{(n-i-1)/2}\Gamma\left[\frac{1}{2}(n-i+1)\right]\right\}=\pi^{m(m-1)/4}\prod_{i=1}^m \Gamma\left[\frac{1}{2}(n-i+1)\right]\cdot2^{mn/2-m} $$

which is written at page 71 in Aspects of Multivariate Statistical Theory by Muirhead.

Answer 1

You omitted the powers of $2$ arising from the first product: $$\prod_{i<j}^m 2^{1/2} = 2^{\binom{m}{2}/2} = 2^{m(m-1)/4}$$

Answer 2

Due to the bijectivity of $\log_2$ (logarithm base $2$), your identity is equivalent to

$$\sum_{i=1}^m \dfrac12(n-i-1) = \dfrac{nm}{2}-m$$

Let us transform its LHS:

$$\dfrac12 \underbrace{\sum_{i=1}^m n}_{nm}- \dfrac12\underbrace{\sum_{i=1}^m( i+1)}_{\tfrac{(m+1)(m+2)}{2}-1} = \dfrac{nm}{2}-m$$

A problem: there is no agreement for the second terms in the LHS and RHS !