Newton's Generalized Binomial Theorem states:$$(a+b)^n=a^n+na^{n-1}b+\dfrac {n(n-1)}{2!}a^{n-2}b^2+\dfrac {n(n-1)(n-2)}{3!}a^{n-3}b^3+\ldots\tag{i}$$
For any number $n$.
And I was going over a book, and it presented the Taylor Expansion for $a^n$ as$$a^n=1+cn+\dfrac {c^2n^2}{2!}+\dfrac {c^3n^3}{3!}+\dfrac {c^4n^4}{4!}+\ldots+\&c\tag{ii}$$
Where $c=(a-1)-\dfrac 12(a-1)^2+\dfrac 13(a-1)^3-\&c=\ln a$. The book also said that $(\text{ii})$ can be proved by using $(\text{i})$. My question:
Question: How would you go about proving $(\text{ii})$ using $(\text i)$?
The book gave this hint:
Let $a^x=\left\{1+(a-1)\right\}^x$ and expand that using $(\text i)$. Collect the coefficients of $x$ to obtain $c$. Assume $c_2,c_3,c_4,\ldots$ as the coefficients of the succeeding powers of $x$ and with this, expand $a^y$ in the same way. Then expand $a^{x+y}$ and equate the coefficients because $a^x\cdot a^y=a^{x+y}$. This will determine $c_2,c_3,c_4,\ldots$.
There are some minor issues with the way you have presented the formulas. Newton's general binomial theorem says that
The result holds for $x = \pm 1$ also but with certain restrictions on $n$. Putting $x = b/a$ and assuming $|b| < |a|$ we can get the series mentioned in your post.
Next we can also write $(1 + x)^{n} = \exp(n\log (1 + x))$ and then using the exponential series we get $$(1 + x)^{n} = 1 + n\log(1 + x) + \frac{(n\log(1 + x))^{2}}{2!} + \cdots$$ Thus we have two series expansions for $(1 + x)^{n}$ and for the series consisting of $\log(1 + x)$ we can see that the coefficient of $n$ is $\log(1 + x)$. The coefficient of the $n$ in the general binomial expansion of $(1 + x)^{n}$ is given by $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots$$ and hence $$\log(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots$$ The above result holds for $-1 < x \leq 1$.
This is the traditional route mentioned in many textbooks. What you are trying to achieve is to get to the exponential series by using binomial theorem. This is very clumsy because you will need to get the expansion of $(\log(1 + x))$ by some other means. It is much simpler to use other approaches to get to exponential series.
But in case you wish to proceed in this manner you can do it with some more effort. Let's assume that we know the expansion of $\log(1 + x)$ by some means and let $$a^{n} = 1 + c_{1}n + \frac{c_{2}}{2!}n^{2} + \cdots$$ then we can write $a^{n} = (1 + (a - 1))^{n}$ and using binomial theorem get $$a^{n} = 1 + nb + \frac{n(n - 1)}{2!}b^{2} + \frac{n(n - 1)(n - 2)}{3!}b^{3} + \cdots$$ where $b = a - 1$. Equating coefficient of $n$ in both series we get $$c_{1} = b - \frac{b^{2}}{2} + \cdots = \log(1 + b) = \log a$$ Next we note that $a^{m + n} = a^{m}a^{n}$ and hence $$1 + c_{1}(m + n) + \frac{c_{2}}{2!}(m + n)^{2} + \cdots = \left(1 + c_{1}m + \frac{c_{2}}{2!}m^{2} + \cdots\right)\left(1 + c_{1}n + \frac{c_{2}}{2!}n^{2} + \cdots\right)$$ From the above we get $$\frac{c_{2}}{2!}(m + n)^{2} = \frac{c_{2}}{2!}m^{2} + \frac{c_{2}}{2!}n^{2} + c_{1}^{2}mn$$ and hence $c_{2} = c_{1}^{2}$ and similarly we can show that $c_{k} = c_{1}^{k}$ so that $c_{k} = (\log a)^{k}$. And hence we get $$a^{n} = 1 + n\log a + \frac{(n\log a)^{2}}{2!} + \cdots$$ Note that the above approach is based on formal manipulation of power series and does not deal with the issues of convergence hence it is not possible to ascertain the range of values of variables for which the formulas hold true. It is best to use rigorous approaches to deal with exponential, logarithmic and binomial series as presented in my blog posts (linked earlier in this answer).