A question about Taylor expansion Is this statement true? Statement: Let $n$ be a positive integer. Consider the Taylor expansion of $\sqrt[n]{1+x}$ to the $k$th order, that is, 
\begin{gather*}
\sqrt[n]{1+x}=\sum_{j=0}^{k}\binom{\frac{1}{n}}{j}x^j+o(x^k), \qquad \text{as $x\to 0$.}
\end{gather*}
Then the Taylor polynomial of $k$th order satisfies 
\begin{gather*}
\left(\sum_{j=0}^{k}\binom{\frac{1}{n}}{j}x^j\right)^n=1+x+\sum_{j=k+1}^{nk}a_jx^j,\tag{1}
\end{gather*}
where $a_j$s are the coefficients which can be determined. 
The interesting part of this statement is that the right hand side of  (1) just a monic polynomial of first order, plus an infinitesimal of very higher-order.
I have checked for $n=2.$ For instance, 
\begin{gather*}
\sqrt{1+x}=1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+O \left( {x}^{5} \right), 
\end{gather*}
and 
\begin{align*}
&\left(1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}\right)^2\\
=&1+x-{\frac {5\,{x}^{7}}{1024}}+{\frac {25\,{x}^{8}}{16384}}+{\frac {7\,{
x}^{6}}{512}}-{\frac {7\,{x}^{5}}{128}}.
\end{align*}
Furthermore, if expand it to $6$th order, then
\begin{gather*}
\sqrt{1+x}=1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+{\frac{7}{256}}{x}^{5}-{\frac{21}{1024}}{x}^{6}+O
 \left( {x}^{7} \right),
\end{gather*}
then we have 
\begin{align*}
&\left(1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+{\frac{7}{256}}{x}^{5}-{\frac{21}{1024}}{x}^{6}\right)^2\\
=&1+x-{\frac {147\,{x}^{11}}{131072}}+{\frac {441\,{x}^{12}}{1048576}}-{
\frac {77\,{x}^{9}}{16384}}+{\frac {77\,{x}^{10}}{32768}}-{\frac {33\,
{x}^{7}}{1024}}+{\frac {165\,{x}^{8}}{16384}}.
\end{align*}
Thus, by examples, it seems that the statement above is true. But I do not know how to prove it. Can you help me?

This is actually a very good exercise in combinatorics and in multiplying polynomials so I will post an answer.
Firstly note that the following identity holds true:
\begin{equation}
1= \sum\limits_{J=0}^{n k} \delta_{j_1+\cdots+j_n,J}
\end{equation}
Now we insert the unity into the left hand side of (1) and we expand the whole thing and we get:
\begin{eqnarray}
lhs(1) = \sum\limits_{J=0}^{n k} x^J \cdot \underbrace{\left( \sum\limits_{j_1=0}^k \sum\limits_{j_2=0}^k  \cdots \sum\limits_{j_n=0}^k \left[\prod\limits_{\xi=1}^n \binom{\frac{1}{n}}{j_\xi}\right] \cdot \delta_{j_1+\cdots+j_n,J}\right)}_{{\mathfrak a}_J}
\end{eqnarray}
Now we evaluate the coefficients $\left\{ {\mathfrak a}_J \right\}_{J=0}^{n k}$ for consecutive values of $J=0,1,2,3,4,\cdots,k,k+1,\cdots,n \cdot k$. We have:
\begin{eqnarray}
{\mathfrak a}_0 &=& 1 \\
{\mathfrak a}_1 &=& n \binom{\frac{1}{n}}{1} = 1 \\
{\mathfrak a}_2 &=& \binom{n}{2} \cdot \binom{\frac{1}{n}}{1}^2 + n \binom{\frac{1}{n}}{2} = 0 \\
{\mathfrak a}_3 &=& \binom{n}{3} \cdot \binom{\frac{1}{n}}{1}^3+\binom{n}{2}(2 \binom{\frac{1}{n}}{2} \binom{\frac{1}{n}}{1}) + n \binom{\frac{1}{n}}{3} = 0 \\
{\mathfrak a}_4 &=& \binom{n}{4} \cdot \binom{\frac{1}{n}}{1}^4+ \binom{n}{3}(3 \binom{\frac{1}{n}}{2} \binom{\frac{1}{n}}{1}^2) +\binom{n}{2} (2 \binom{\frac{1}{n}}{3} \binom{\frac{1}{n}}{1} + \binom{\frac{1}{n}}{2}^2)+n \binom{\frac{1}{n}}{4} =0
\end{eqnarray}
Let us analyze the last line above. We can achieve $J=4$ in four different ways(from the left to the right), firstly by taking four distinct $j$-indices equal to unity, secondly by taking one $j$-index equal to two and two other ones equal to one, thirdly by taking one index equal to three and another one to one or two distinct indices being equal to two and fourthly by taking exactly one index equal to four. The corresponding numbers of ways for achieving that are given by the respective binomial factors which stand in front of each term in the right hand side of the last line.
Now, clearly for bigger values of $J$ nothing much will change except that we only have to enumerate more possibilities and the expressions become lengthy.  The only interesting thing happens when $J$ hits the value $J=k+1$. Then clearly the very last term on the right hand side is missing and therefore:
\begin{eqnarray}
{\mathfrak a}_{k+1} &=& -n \binom{\frac{1}{n}}{k+1} 
\end{eqnarray}
which gives ${\mathfrak a}_5 = - 7/128$ for  $(n,k)=(2,4)$ and ${\mathfrak a}_7 = - 33/1024$ for $(n,k)=(2,6)$ as in the question above. The higher order coefficients are more complicated but it is clear how they can be extracted. As $J$ increases there will be more and more terms missing on the very right of the right hand side. One needs to extract those terms and carefuly sum them up to get the coefficient in question.
