What we can infer from equating Taylor series for $(1+x)^n$ and binomial theorem for $(1+x)^n$? $(1+x)^n=1+\frac{(n)ln(1+x)}{1!}+\frac{(n^2)(ln(1+x))^2}{2!}+\frac{(n^3)(ln(1+x))^3}{3!}+\cdots$
$(1+x)^n=1+nx+(^nC_2){x^2}+(^nC_3){x^3+\cdots}$
Here, equating both the series doesn't make any sense of progress to get any sort of result since the former one contains natural logs of $(1+x)$ and the later one has factorial values attached to the variable $x$. So, comparing the coefficients looks like a tedious and cumbersome task. But I was still wondering if we can extract any result from the comparison of the two forms? 

As said in comment, what you did wrong,was to consider that $$(1+x)^n= e^{n\, \ln(1+x)}=e^y$$ and you did apply the expansion of $$e^y=1+y+ \frac{y^2}{2!}+ \frac{y^3}{3!}+\cdots$$ and making $y=n\ln(1+x)$ $$(1+x)^n=1+\frac{n\ln(1+x)}{1!}+\frac{n^2 \ln^2(1+x)}{2!}+\frac{n^3 \ln^3(1+x)}{3!}+\cdots\tag 1$$ which is your first expression.
However, you can still recover using Taylor expansion of $\ln(1+x)$ since $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ $$\ln^2(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)=x^2-x^3+O\left(x^4\right)$$ $$\ln^3(1+x)=x^3+O\left(x^4\right)$$ Replacing in $(1)$, you then get $$(1+x)^n=1+n x+\left(\frac{n^2}{2}-\frac{n}{2}\right)
   x^2+\left(\frac{n^3}{6}-\frac{n^2}{2}+\frac{n}{3}\right) x^3+O\left(x^4\right)$$ which, after simplifications, leads to $$(1+x)^n=1+n x+\frac{1}{2} n(n-1)  x^2+\frac{1}{6} n(n-1)(n-2)  x^3+O\left(x^4\right)$$ which is your second expression.
You took a long way $\cdots$ but it works !
