Why is the standard inner product on F^n equal to this?

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In the textbook that I'm using the standard inner product is defined as

$$\langle x,y\rangle = \sum_{i=1}^{n}a_{i}\overline{b_{i}}$$

where $x=(a_{1}, a_{2}, {...}, a_{n})$ and $y=(b_{1}, b_{2}, {...}, b_{n})$ and the bar denotes complex conjugation.

In one the practice problems I was doing they used the fact that

$\langle x,y\rangle = y^{*}x$

where $y^{*}$ denotes the conjugate transpose of y. I'm having trouble understanding why the definition for the standard inner product is the same as $y^{*}x$.

I tried doing an example where $x = (i,i)$ and $y = (-i,-i)$

For $$\langle x,y\rangle = i*i+i*i=-2$$

On the other hand

$y^{*}x=\begin{pmatrix} i\\i\\ \end{pmatrix} \begin{pmatrix} i & i\\ \end{pmatrix} = \begin{pmatrix} -1 & -1\\ -1 & -1\\ \end{pmatrix} $

I was wondering why I am getting $\langle x,y\rangle \ne y^{*}x$