If the elements $y_1,y_2,\dots,y_r \in E$ are linearly independent over $F$, show that $y_1^p,y_2^p,\dots,y_r^p$ are linearly independent over $F$. 
Let $E$ be a finite field extension of a field $F$ of a prime characteristic $p$, and assume that $E=F(E^p)$. If the elements $y_1,y_2,\dots,y_r \in E$ are linearly independent over $F$, show that $y_1^p,y_2^p,\dots,y_r^p$ are linearly independent over $F$.

Take note, (I) there is one related exercise showing the following result, which I already know.

Let $E$ be a finite field extension of a field $F$ of a prime characteristic $p$, and assume that $K=F(E^p)$ be the subfield of $E$ obtained by adjoining the $p$th powers of all elements in $E$. Show that $F(E^p)$ consists of all finite linear combinations of elements in $E^p$ with coefficients in $F$.

(II) There are two similar posts found on MSE, but i am not satisfied with them.
If $K=F(K^p)$ is a finite extension and $\{a_1,\ldots,a_n\} \subset K$ linearly independent then so is $\{{a_1}^p,\ldots,{a_n}^p \}$
Suppose $K/F $ is finite with $K = F (K^p) $. If a finite set $S $ is linearly independent, prove that so is $S^p $.
I want a more detailed answer by showing that if $\sum_{i=0}^ra_iy_i^p=0$, then $a_i=0$ for all $i\in\{1,2,\dots,r\}$.
 A: The following is a detailed proof following the hint posted as an answer by
@user26857. We will use the
following fact:

Lemma 1. Let $K$ be a field. Let $L$ be a $K$-algebra that is
finite-dimensional as a $K$-vector space. Let $R$ be a $K$-subalgebra of $L$
that is an integral domain. Then, $R$ is a field.

Lemma 1 is the particular case of the Theorem 1 from
https://math.stackexchange.com/a/3161397/ when  $g$ is an inclusion map. (Of
course, this particular case is easily seen to be equivalent to the general case.)
Let us now solve your question.
Assume that $y_{1},y_{2},\ldots,y_{r}$ are $r$ elements of $E$ that are
linearly independent over $F$. We must show that the $r$ elements $y_{1}
^{p},y_{2}^{p},\ldots,y_{r}^{p}$ of $E$ are linearly independent over $F$ as well.
The list $\left(  y_{1},y_{2},\ldots,y_{r}\right)  $ of vectors in $E$ is
linearly independent over $F$; thus, we can extend this list to a basis
$\left(  y_{1},y_{2},\ldots,y_{n}\right)  $ of the $F$-vector space $E$ (since
$E$ is a finite-dimensional $F$-vector space). Consider such a basis $\left(
y_{1},y_{2},\ldots,y_{n}\right)  $. Since $\left(  y_{1},y_{2},\ldots
,y_{n}\right)  $ is a basis of $E$, there is a unique endomorphism $\phi$ of
the $F$-vector space $E$ that sends $y_{1},y_{2},\ldots,y_{n}$ to $y_{1}
^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Consider this $\phi$.
The image $\phi\left(  E\right)  $ of $E$ is an $F$-vector subspace of $E$
(since $\phi$ is $F$-linear), but it has further properties. To wit, let us
first show that $E^{p}\subseteq\phi\left(  E\right)  $.
Indeed, let $w\in E^{p}$. Then, $w=e^{p}$ for some $e\in E$. Consider this
$e$. Then, $e=\sum_{i=1}^{n}a_{i}y_{i}$ for some $a_{1},a_{2},\ldots,a_{n}\in
F$ (since $\left(  y_{1},y_{2},\ldots,y_{n}\right)  $ is a basis of the
$F$-vector space $E$). Consider these $a_{1},a_{2},\ldots,a_{n}$. Now,
\begin{align*}
w  & =e^{p}=\left(  \sum_{i=1}^{n}a_{i}y_{i}\right)  ^{p}\qquad\left(
\text{since }e=\sum_{i=1}^{n}a_{i}y_{i}\right)  \\
& =\sum_{i=1}^{n}a_{i}^{p}y_{i}^{p}
\end{align*}
(since we are in characteristic $p>0$, so that the map $E\rightarrow
E,\ z\mapsto z^{p}$ is a ring endomorphism of $E$). On the other hand, the
definition of $\phi$ yields
\begin{align*}
\phi\left(  \sum_{i=1}^{n}a_{i}^{p}y_{i}\right)  =\sum_{i=1}^{n}a_{i}^{p}
y_{i}^{p}
\end{align*}
(since $a_{1},a_{2},\ldots,a_{n}$ are scalars in $F$). Comparing these two
equalities, we find $w=\phi\left(  \sum_{i=1}^{n}a_{i}^{p}y_{i}\right)
\in\phi\left(  E\right)  $.
Forget that we fixed $w$. We thus have shown that $w\in\phi\left(  E\right)  $
for each $w\in E^{p}$. In other words, $E^{p}\subseteq\phi\left(  E\right)  $.
Hence, $1\in E^{p}\subseteq\phi\left(  E\right)  $, so that $F=F\cdot
1\subseteq\phi\left(  E\right)  $ (since $\phi\left(  E\right)  $ is an
$F$-vector subspace of $E$ and since $1\in\phi\left(  E\right)  $).
Next, we shall show that the set $\phi\left(  E\right)  $ is closed under
multiplication. Indeed, $\phi\left(  E\right)  $ is the image of the
$F$-linear map $\phi$, which sends the basis elements $y_{1},y_{2}
,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Thus,
$\phi\left(  E\right)  $ is the $F$-linear span of the elements $y_{1}
^{p},y_{2}^{p},\ldots,y_{n}^{p}$. Hence, in order to show that $\phi\left(
E\right)  $ is closed under multiplication, it will suffice to prove that the
pairwise products $y_{i}^{p}y_{j}^{p}$ of the latter elements $y_{1}^{p}
,y_{2}^{p},\ldots,y_{n}^{p}$ belong to $\phi\left(  E\right)  $. But this is
easy: For any $i,j\in\left\{  1,2,\ldots,n\right\}  $, we have $y_{i}^{p}
y_{j}^{p}=\left(  y_{i}y_{j}\right)  ^{p}\in E^{p}\subseteq\phi\left(
E\right)  $. Thus, we have shown that $\phi\left(  E\right)  $ is closed under
multiplication. Since $\phi\left(  E\right)  $ is furthermore an $F$-vector
subspace of $E$ and contains $1$ (since $1\in\phi\left(  E\right)  $), we thus
conclude that $\phi\left(  E\right)  $ is an $F$-subalgebra of $E$. In
particular, $\phi\left(  E\right)  $ is a subring of the field $E$, and thus
is an integral domain (since any subring of a field is an integral domain).
Hence, Lemma 1 (applied to $K=F$ and $L=E$ and $R=\phi\left(  E\right)  $)
yields that $\phi\left(  E\right)  $ is a field. Therefore, $\phi\left(
E\right)  $ is a subfield of $E$ that contains both $F$ and $E^{p}$ as
subfields (since $F\subseteq\phi\left(  E\right)  $ and $E^{p}\subseteq
\phi\left(  E\right)  $).
However, $F\left(  E^{p}\right)  $ is the smallest such subfield (by the
definition of $F\left(  E^{p}\right)  $). Hence, $F\left(  E^{p}\right)
\subseteq\phi\left(  E\right)  $. Now, recall that $E=F\left(  E^{p}\right)
$. Hence, $E=F\left(  E^{p}\right)  \subseteq\phi\left(  E\right)  $.
Combining this with $\phi\left(  E\right)  \subseteq E$, we obtain
$E=\phi\left(  E\right)  $. In other words, the map $\phi$ is surjective.
However, a surjective endomorphism of a finite-dimensional $F$-vector space
must necessarily be injective. Thus, $\phi$ is injective (since $\phi$ is a
surjective endomorphism of the finite-dimensional $F$-vector space $E$). Since
$\phi$ is the $F$-linear map that sends the basis vectors $y_{1},y_{2}
,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, this entails that
the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$ are $F$-linearly
independent. Hence, the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{r}^{p}$ are
$F$-linearly independent (since $n\geq r$). Qed.
