\(n\) | \(\hspace{10mm} 2^n\) | Concatenation of the powered numbers | Divisibility checked |

0 | \(\hspace{5mm} 2^0=1\) | \(\hspace{25mm} 1\) | \(\hspace{8mm} 1\, \big|\, 1\) |

1 | \(\hspace{5mm} 2^1=2\) | \(\hspace{25mm} 12\) | \(\hspace{8mm} 2\, \big|\, 12\) |

2 | \(\hspace{5mm} 2^2=4\) | \(\hspace{25mm} 124\) | \(\hspace{8mm} 4\, \big|\, 124\) |

3 | \(\hspace{5mm} 2^3=8\) | \(\hspace{25mm} 1248\) | \(\hspace{8mm} 8\, \big|\, 1248\) |

4 | \(\hspace{5mm} 2^4=16\) | \(\hspace{25mm} 124816\) | \(\hspace{8mm} 16\, \big|\, 124816\) |

As we get greater and greater numbers in column 3 of the table by concatenation (i.e. 12481632, 1248163264, ...) for \(n>4,\) will the divisibility in the last column still hold?

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