Let ${a_n} = \prod\limits_{j = 1}^n {\gcd (j,n)} .$ Comparing OEIS A067911 and A170911 suggests that there are integers $b_n$ such that the $n-$th partial sum of the product $\prod\limits_{k = 0}^\infty {\left( {1 + \frac{{{b_k}}}{{{a_k}}}{x^k}} \right)} $ coincides with the $n-$th partial sum $\sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} $ of $exp(x).$ Is there a proof of this fact in the literature?
Edit:
The first terms of the sequence $a_n$ are $1, 1, 2, 3, 8, 5, 72, 7, 128, 81, 800, 11, 41472, 13, \dots.$
The first terms of the sequence $b_n$ are $0, 1, 1,-1, 3, -1, 13, -1, 27, -8, 91, -1,3639, -1,\dots.$ $$(1+x)(1+\frac{x^2}{2})=1+\frac{x}{1!}+ \frac{x^2}{2!}+\dots,$$ $$(1+x)(1+\frac{x^2}{2})(1-\frac{x^3}{3})=1+\frac{x}{1!}+ \frac{x^2}{2!} + \frac{x^3}{3!}+\dots,$$ $$(1+x)(1+\frac{x^2}{2})(1-\frac{x^3}{3})(1+\frac{3x^4}{8})=1+\frac{x}{1!}+ \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\dots.$$ In the mean-time I found some papers which lead to a proof by using the power series expansion of the logarithm. On the other hand by comparing coefficients of the partial products we get rational numbers $b_n.$ Is there a direct method to see that these numbers are in fact integers?