Can we find this integral involving the floor function?

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For a real number \(x\) we denote by \([x]\) the largest integer less than or equal to \(x\).

Let \(n\) be a natural number. The integral \[\int_0^n [2^x]\, dx\] equals

1. \(\log_2 ((2^n-1)!)\);

2. \(n2^n-\log_2 ((2^n)!)\);

3. \(n2^n\);

4. \(\log_2 ((2^n)!)\),

where \(k! = 1\times 2 \times 3 \times \cdots \times k\) for a positive integer \(k\).