$$T= \prod_{i=2}^{N} \left ( i^3 - 1 \right ) ~~ \text{Find T in terms of N}$$
This seems like a super easy task but somehow I cannot figure it out. My try was to convert it to a sum of logs, denote
$$ \prod_{i=2}^{N} \left ( i^3 - 1 \right ) = T \\ \lg \left (\prod_{i=2}^{N} \left ( i^3 - 1 \right ) \right) = \lg (T) \\ \sum_{i=2}^{N} \lg(i^3 -1) = \lg(T)$$
But here I am stuck as $i^3 -1$ does not factor nicely, so it is a dead-end.. (I am aware that $i^2 -1$ would word nice actually, but here it does not work).
I know that:
$$\prod_{i=2}^{N} i^3 = 2^3 \cdot 3^3 \dots \cdot N^3 = \left (N! \right)^3$$
And so I just need to "add $+1$" to each term and then find the value, from a property of multiplication:
$$ (A-1)\cdot B = A \cdot B - B$$
And so, we add $+1$ to the $2^3 -1$ term, then subtract the rest:
$$ T = 2^3 \cdot (3^3 -1) \dots \cdot (N^3-1) \\ ~~ - (3^3 -1) \dots (N^3-1) $$
We do the same for $3^3-1$ and so on... and we get this 'triangular' shape:
$$ T = 2^3 \cdot 3^3 \dots \cdot N^3 \\ ~~ - (3^3 -1) \dots (N^3-1) \\ ~~~ - 2^3(4^3-1)\dots(N^3-1) \\ ~~~~ \vdots$$
But again, this seems like a dead end, and it is hard to evaluate by itself!
Any help would be appreciated! Thank you.
This is sequence $A158620$ in $OEIS$. There is an approximation Vaclav Kotesovec gave in $2015$ which is $$T_n\sim \frac{2\sqrt{2 \pi }}{3} \cosh \left(\frac{\sqrt{3\pi} }{2} \right)\,n^{3 n+\frac{3}{2}}\, e^{-3 n} $$
Generating the values for $2 \leq n \leq 100$ (remember that $T_{100}\sim 6.58\times 10^{472}$ !), I made a quick and dirty linear regression $$\log(T_n)=a+b\, n+c \log(n)+d\, n \log(n)$$
The results are reported below $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +2.77075 & 0.00562 & \{+2.75958,+2.78191\} \\ b & -2.98305 & 0.00105 & \{-2.98513,-2.98097\} \\ c & +1.37264 & 0.00499 & \{+1.36274,+1.38254\} \\ d & +2.99710 & 0.00019 & \{+2.99672,+2.99747\} \\ \end{array}$$
Forcing $d=3$ now shows that $b$ is extremely close to $-3$. So using $b=-3$ and $d=3$ leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 2.64594 & 0.00727 & \{2.63151,2.66037\} \\ c & 1.47615 & 0.00193 & \{1.47232,1.47997\} \end{array}$$ Now $c\sim \frac 32$ and this leads closer and closer to Vaclav Kotesovec's approximation.
Edit
Using a CAS $$T_n=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{3 \pi }\Gamma (n) \Gamma \left(n+\frac{3+i \sqrt{3}}{2}\right) \Gamma \left(n+\frac{3-i \sqrt{3}}{2}\right)$$ where appears, as natural, the product of three gamma functions.
Using now Stirling approximation (any other could be possible) of the gamma function, we have $$T_n=\frac{2\sqrt{2 \pi }}{3} \cosh \left(\frac{\sqrt{3\pi} }{2} \right)\,n^{3 n+\frac{3}{2}}\, e^{-3 n+f(n)} $$ where $$f(n)=\frac{1}{4 n}+\frac{1}{2 n^2}-\frac{61}{120 n^3}+\frac{1}{4 n^4}+\frac{43}{420 n^5}-\frac{1}{3 n^6}+\frac{139}{560 n^7}+O\left(\frac{1}{n^8}\right)$$
This truncated expression gives $T_2=6.98683$ (instead of $7$) and $T_3=181.979$ (instead of $182$).
Now, since $$U_n=\prod_{i=2}^n i^3=\Big[\Gamma(n+1)\Big]^3$$ $$\prod_{i=2}^{n} \left ( 1 - \frac{1}{i^3} \right )=\frac {T_n}{U_n}\sim\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{3 \pi } \exp\Bigg[\frac{1}{2 n^2}-\frac{1}{2 n^3}+\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right) \Bigg]$$