Your original statement is not true (please read the whole answer). But the following statement is correct:
$$\left \lfloor {\log_2 n} \right \rfloor = \left \lfloor {\log_2 \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor + 1\tag{1}$$
Proof:
Every number $n$ can be placed between two cosecutive powers of 2. In other words, there exists $k$ such that:
$$2^k\le n\le2^{k+1}-1\tag{1}$$
Obviously:
$$k\le\log_2n<k+1$$
$$k=\lfloor\log_2n\rfloor\tag{2}$$
On the other side from (1):
$$\frac{2^k-1}2 \le \frac{n-1}{2} \le \frac{2^{k+1}-2}2$$
$$2^{k-1}-\frac12 \le \frac{n-1}{2} \le 2^k-1$$
$$\lceil 2^{k-1}-\frac12\rceil \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$$2^{k-1} \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$${k-1} \le \log_2\lceil\frac{n-1}{2}\rceil \le \log_2 (2^k-1)<k$$
$${k-1} =\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor$$
$$k=\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1\tag{3}$$
By comparing (2) and (3) you get:
$$\lfloor\log_2n\rfloor = \lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1$$
...which completes the proof.
You can easily prove that the original statement is not true. You are basically saying that the function:
$$f(n)=\left \lfloor {\log n} \right \rfloor - \left \lfloor {\log \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor - 1$$
...is equal to zero for all values of $n$.
This is not true if "$\log$" stands for logartihm with base 10:
This is also not true if "$\log$" stans for natural logarithm "$\ln$":
If you don't trust these plots, calculate the value for $n=45$ and in both cases the result is -1, not 0.
Best Answer
You need to replace the ceiling function with the floor function. Let $N=lcm(1,\dots,n)$. If $p$ is a prime and $k$ and integer bigger or equal than $1$, then you have:
1) if $p^k\le n$, then $p^k|N$ by definition of $N$;
2) if $p^k|N$, then there must be an integer $1\le m\le n$ such that $p^k|m$, by definition of $lcm$ (and prime numbers). Therefore $p^k\le n$.
It follows that for any prime $p$, $p^k|N$ if and only if $p^k\le n$, which is equivalent to $k\le \lfloor \log_p n \rfloor$. Therefore the exponent of the power of $p$ in the prime factorization of $N$ is $\lfloor \log_p n \rfloor$ as claimed.