Prove $2^{lceil lg (n-1) rceil} ge n-1$

Note that $2^{lg(n-1)} = n-1$ for all $n > 1$. (Here, we use the base $2$ logarithm $lg$ instead of $ln$, as using the natural logarithm would lead the inequality being false for all sufficiently large $n$. This usage is probably what the problem implied.) Then, since $lceil xrceil geq x$ for all $x$, and $2^{x}$ is an increasing function for all $x$, we have:

$$2^{lceillg(n-1)rceil}geq 2^{lg(n-1)} = n-1 blacksquare$$