If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$

  • A

    $0$

  • B

    $1$

  • C

    $n$

  • D

    $2n$

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