Does a function $f$ exists such that: $f(n-k) \ne \Theta(f(n))$ for some constant $k\geq1$?

I have encountered the following question in my homework assignment in Data Structures course:

"Does a function $$f$$ exists such that: $$f(n-k) \ne \Theta(f(n))$$ for some constant $$k\geq1$$ ?"

I think no such function $$f$$ exists, but I do not know how to prove it (or give a counter-example if one exists).

• For example, let $f(n)=n$ if $n$ is a square and $f(n)=1$ otherwise. For every $k\ge1$, $f(n-k) \ne \Theta(f(n))$ – John L. Aug 25 '20 at 18:52

An example is $$f(n) = 2 ^ {2^n}$$. Now, $$f(n-1) = 2 ^ {2^{n-1}}$$ and we have $$\frac{f(n-1)}{f(n)} = \frac{1}{2^{2^n - 2^{n-1}}} = \frac{1}{2^{2^{n-1}}}$$. Hence, $$f(n-1) \not \in \Theta(f(n))$$. In this example, $$k = 1$$.
Counterexample: $$f(n)=n!$$ As $$f(n)$$ is $$n$$ times bigger than $$f(n-1)$$, it is clear that $$f(n-1) \neq \Theta(f(n))$$.