Product of n terms of a geometric progression

The general term of this progression is $a_n=\sqrt{7}\cdot r^{n-1}$ since we don't know the value of the ratio, but we do know that the first term is $\sqrt{7}$.

On the other hand, the product of the first six terms is $\sqrt{7^{21}}$, and if we apply the formula, we have:

$$P_6=\sqrt{(a_1\cdot a_6)^6}=\sqrt{(\sqrt{7}\cdot\sqrt{7}\cdot r^5)^6}= \sqrt{7^6\cdot r^{30}}= 7^3\cdot r^{15}$$

So,

$$7^3\cdot r^{15} = \sqrt{7^{21}} \Rightarrow r^{15}=\sqrt{7^{15}} \Rightarrow r^{15}=(\sqrt{7})^{15} \Rightarrow r=\sqrt{7}$$

Therefore, the general term of the succession is: $$a_n=\sqrt{7^n}$$ So, $$a_1=\sqrt{7}, a_2=7, a_3=7\sqrt{7}, a_4=49, a_5=49\sqrt{7}$$

The general term of the succession with the first term $a_1=1$ and ratio $r=\dfrac{a_2}{a_1}=\dfrac{0.1}{1}=0.1=\dfrac{1}{10}$, is

$$a_n=\dfrac{1}{10^{n-1}}$$

We want to find a natural $m$ in such a way that the product of the $m$ first terms of the succession is $10^{-45}$, that is to say, that

$$P_m=\prod_{n=1}^m \dfrac{1}{10^{n-1}} = 1^{10}\cdot 10^{-45}$$

but we know that:

$$P_m=\sqrt{(a_1\cdot a_m)^m}=\sqrt{ \Big(1\cdot \dfrac{1}{10^{m-1}}\Big)^m}$$

And by comparing both expressions, we have:

$$10^{-45}= \sqrt{ \Big(\dfrac{1}{10^{m-1}}\Big)^m}$$

And by isolating the variable of this rational equation:

$$\Big(\dfrac{1}{10^{m-1}}\Big)^{\frac{m}{2}}=10^{-45} \Rightarrow \dfrac{1}{10^{\frac{m}{2}(m-1)}}=\dfrac{1}{10^{45}} \Rightarrow$$

$$10^{\frac{m(m-1)}{2}}=10^{45} \Rightarrow \dfrac{m^2-m}{2}=45 \Rightarrow$$

$$m^2-m-90=0$$

So we only have to solve this equation of the second grade:

$$m^2-m-90=0 \Rightarrow m=\{10,-9\}$$

We know that $m$ must be a positive integer, so the solution is $m=10$.