Find what a Infinite product number approaches

Question

Find what the following number approaches: $$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$

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What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$ Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$)

What should I do next?

Note: Please use junior high school math if possible

Answer 1

You have

$$\left(1+\frac{1}{5^1}\right)\left(1+\frac{1}{5^2}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}+\frac{1}{5^8}+\cdots$$

$$= \frac{5}{5-1} \\= 1+\frac{1}{4}$$

Answer 2

Put $x=\frac{1}{5}$ then we have $$\lim_{n \to \infty}(1+x)(1+x^2)(1+x^4)(1+x^8)..(1+x^{2^n})$$ now multiply $1-x$ in numerator and denominator and use $(1-x^{2^k})(1+x^{2^k})=1-x^{2^{k+1}}$ repeatedly and see the magic!