Note that I am aware of these related links:
How to prove Gauss’s Multiplication Formula?
Ahlfors "Prove the formula of Gauss"
(what I want is getting help regarding my try using induction if that is possible)
Prove the formula of Gauss: $$ (2\pi)^{\frac{n-1}{2}} \Gamma(z) = n^{z-\frac{1}{2}} \Gamma\left(\frac{z}{n}\right) \Gamma\left(\frac{z+1}{n}\right) \cdots \Gamma\left(\frac{z+n-1}{n}\right). $$
Proof: By induction.
For (n = 2): $$ (2\pi)^{\frac{1}{2}} \Gamma(z) = 2^{z-\frac{1}{2}} \Gamma\left(\frac{z}{2}\right) \Gamma\left(\frac{z+1}{2}\right), $$ which is true by the property (which is proved in Ahlfors' book page 200): $$ \sqrt{\pi} \, \Gamma(2z) = 2^{2z-1} \Gamma(z) \Gamma\left(z+\frac{1}{2}\right). $$
Replacing $z$ by $\frac{z}{2}$ $$ (\pi)^{\frac{1}{2}} \Gamma(z) = 2^{z-1} \Gamma\left(\frac{z}{2}\right) \Gamma\left(\frac{z+1}{2}\right). $$
Multiply both sides by $\sqrt{2}$: $$ \sqrt{2} \cdot (\pi)^{\frac{1}{2}} \Gamma(z) = \sqrt{2} \cdot 2^{z-1} \Gamma\left(\frac{z}{2}\right) \Gamma\left(\frac{z+1}{2}\right), $$ which simplifies to: $$ (2\pi)^{\frac{1}{2}} \Gamma(z) = 2^{z-\frac{1}{2}} \Gamma\left(\frac{z}{2}\right) \Gamma\left(\frac{z+1}{2}\right). $$
Now, assume the statement is true for $n = k$: $$ (2\pi)^{\frac{k-1}{2}} \Gamma(z) = k^{z-\frac{1}{2}} \Gamma\left(\frac{z}{k}\right) \Gamma\left(\frac{z+1}{k}\right) \cdots \Gamma\left(\frac{z+k-1}{k}\right). $$
We want to prove it for $n = k+1$, which means: $$ (2\pi)^{\frac{k}{2}} \Gamma(z) = (k+1)^{z-\frac{1}{2}} \Gamma\left(\frac{z}{k+1}\right) \Gamma\left(\frac{z+1}{k+1}\right) \cdots \Gamma\left(\frac{z+k}{k+1}\right). $$
To prove this, start with: $$ (2\pi)^{\frac{k}{2}} \Gamma(z) = \sqrt{2\pi} \cdot (2\pi)^{\frac{k-1}{2}} \Gamma(z). $$
By the induction hypothesis: $$ (2\pi)^{\frac{k}{2}} \Gamma(z) = \sqrt{2\pi} \cdot k^{z-\frac{1}{2}} \Gamma\left(\frac{z}{k}\right) \Gamma\left(\frac{z+1}{k}\right) \cdots \Gamma\left(\frac{z+k-1}{k}\right). $$
But, I don't how to reach the required result, is this approach impossible to solve it?
