Proof of ⌊Log2[n]⌋ + 1 = ⌈Log2[n+1]⌉

n is a positive integer here

⌊Log2[n]⌋≤ Log2[n]<⌊Log2[n]⌋+1

(2^(⌊Log2[n]⌋))≤ (2^(Log2[n]))<(2^(⌊Log2[n]⌋+1))

(2^(⌊Log2[n]⌋))≤ n<(2^(⌊Log2[n]⌋+1))                  (1)

(2^(⌊Log2[n]⌋))< n+1≤ (2^(⌊Log2[n]⌋+1))           (2)

[the right-hand inequality ≤ in (2) follows from the right-hand inequality < in (1)  because n and (2^(⌊Log2[n]⌋+1))are an integers]

Log2[(2^(⌊Log2[n]⌋))]< Log2[n+1]≤ Log2[(2^(⌊Log2[n]⌋+1))]

⌊Log2[n]⌋< Log2[n+1]≤ ⌊Log2[n]⌋+1

⌈Log2[n+1]⌉= ⌊Log2[n]⌋+1

A variant with k

Let k =⌊Log2[n]⌋

⌊Log2[n]⌋≤ Log2[n]<⌊Log2[n]⌋+1

k≤ Log2[n]<k+1

(2^k)≤ (2^(Log2[n]))<(2^(k+1))

(2^k)≤ n<(2^(k+1))                                        (3)

(2^k)< n+1≤ (2^(k+1))                                 (4)

[the right-hand inequality ≤ in (4) follows from the right-hand inequality < in (3)  because n and (2^(k+1))are an integers]

Log2[(2^k)]< Log2[n+1]≤ Log2[(2^(k+1))]

k< Log2[n+1]≤ k+1

⌈Log2[n+1]⌉= k+1

⌈Log2[n+1]⌉= ⌊Log2[n]⌋+1

Optional Note

For any positive real number x such that 
(2^i)-1<x<(2^i)
for some integer i,
the equality

⌈Log2[x+1]⌉= ⌊Log2[x]⌋+1

is false.

For all other positive real numbers x, the equality

⌈Log2[x+1]⌉= ⌊Log2[x]⌋+1

is true.

Plot[Tooltip[{Floor[Log2[x]]+1+0.04,Ceiling[Log2[x+1]]}],{x,0,16},PlotTheme->"Classic"]