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Suppose $F$ be a field of characteristic $p$,let $E$ be an extension of $F$ such that $E$ contains all $p^r$,$r \geq0$ roots of elements of $F$ then $E$ is perfect.

Perfect Field: A field $F$ is called Perfect if the map $F \to F$ defined as $a \to a^p$ is bijective.

I can see it intuitively but unable to prove it precisely.

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    Well, what do you do in order to show that something is bijective? – 2017-02-06
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    @Starfall: Basically i need to just show that map is surjective. – 2017-02-06
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    ...and how do you do that? Can you write out, explicitly, what it means for this map to be surjective? – 2017-02-06
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    @Starfall: it means that every element of $E$ is pth root of some element of $E$ ? – 2017-02-06
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    @MathLover Don't worry, I am missing something here, too. I messed up between $\;E,\,F\;$ . yet I still think the claim is true. I'm deleting the other comment to avoid misunderstandings. – 2017-02-06

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$ F = \mathbb F_p $, $ E = \mathbb F_p(T) $ is a counterexample.

Edit: It appears that the OP made a mistake when asking the question, and they wanted to specify that $ E $ is actually generated over $ F $ by the $ p^r $-th roots of the elements of $ F $. In this case, the claim is indeed true: the map $ x \to x^p $ is surjective, because given any $ y \in E $, we may write

$$ c_1 z_1 + c_2 z_2 + \ldots + c_n z_n $$

with $ c_i \in F $ and $ z_i \in E $, such that each $ z_i $ is a $ p^{r_i} $-th root of an element of $ F $. Then,

$$ (c_1^{1/p} z_1^{1/p} + c_2^{1/p} z_2^{1/p} + \ldots + c_n^{1/p} z_n^{1/p})^p = c_1 z_1 + c_2 z_2 + \ldots + c_n z_n $$

and the base of the exponentiation on the LHS lies in $ E $, as

$$ (c_i^{1/p} z_i^{1/p})^{p^{r_i + 1}} = (c_i z_i)^{p^{r_i}} \in F $$

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    Nice. +1 The assumption must be stronger, like $\;E\;$ containing all $\;p-th\;$ roots of elements in $\;E\;$ . I wonder though whether this would make $\;E\;$ algebraically closed... – 2017-02-06
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    @DonAntonio It does not, and it's fairly easy to see why. Simply considering the compositum of all finite extensions of $ \mathbb F_p $ of degree a power of $ p $ does the trick. – 2017-02-06
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    @Starfall,Actually i was reading a note to construct Perfect closure and there it was said that just attach all the $p^r$ th roots of elements of $F$ so claim is not true? – 2017-02-06
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    @MathLover That is not the claim you wrote in the question - you did not specify that $ E $ is the extension one gets after adjoining those roots, you simply said that it *contains* those roots. With your additional condition, the claim is indeed true. – 2017-02-06
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    Ah i see.How the claim is true in this case? – 2017-02-06
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    @Starfall Of course. Once again, I misunderstood the condition and thought of the $\;p\,-$ th roots as $\;p\;$ ranging over all natural numbers. Gees, I'm on fire today... – 2017-02-06
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    @Starfall: Thank you! – 2017-02-06
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    Why should $E$ be the *vector space* generated by $p^r$-th roots? – 2017-02-06
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    @egreg The set of all finite $ F $-linear combinations of the $ p^r $-th roots of elements in $ F $ forms a field, and it's clearly the smallest extension of $ F $ containing all of those roots. – 2017-02-06
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    @Starfall It's not at all obvious that $(c_1 z_1 + c_2 z_2 + \dots + c_n z_n)^{-1}$ is a linear combination of $p^r$-th roots. – 2017-02-06
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    @egreg The inverse of an algebraic number is a polynomial function of that algebraic number. – 2017-02-06
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    @Starfall That's what I asked you to observe. – 2017-02-06
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    @egreg I fail to see how this is "not obvious". – 2017-02-06