If I have $x$ bits I can represent the numbers from $0$ to $2^{x}-1=n-1$.
Let $x=2$.
$2^{x}=4:00, 01, 10, 11=3=4-1$.

On the other hand, If I want to represent up to $n-1$ I need $x=\lceil lg(n)\rceil$ bits.
If $n-1=4$ then $n=5$ and $\lceil lg(5)\rceil=\lceil ln(5)/ln(2)\rceil=\lceil 2.32...\rceil=3$.
Indeed $4$ is $100$ and needs $3$ bits.

Let's get to the picture.
Up to $n$ I need $\lceil lg(n+1)\rceil$ bits.
I guess that $\lceil lg(n+1)\rceil=lg(n+t)$ (integer) from some $t\geq 1$.
$lg(n+t)=clg(n)$ is solved by $c=lg(n+t)/lg(n)\geq 1$.

[we restrict $c$ to be a constant so that the word size does not grow arbitrarily]

I don't understand this quote.
In my math $c$ is the only solution.