Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:

Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.

It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!

Let $s_1:=x+\dfrac1x$. Then
$$x=\frac{s_1\pm\sqrt{s_1^2-4}}2$$ and
$$s_n=\left(\frac{s_1+\sqrt{s_1^2-4}}2\right)^n+\left(\frac{s_1-\sqrt{s_1^2-4}}2\right)^n.$$
Then by the Binomial development, after cancellation of the odd terms
$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\binom n{2k} s_1^{n-2k}\left(s_1^2-4\right)^{k}.$$
Then, developing again,
$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\sum_{j=0}^k\binom n{2k}\binom kj(-4)^{k-j} s_1^{n-2k+2j}$$
and you can regroup the terms by equal powers of $s_1$.
