# Are NP problems lower bounded by exponential order of growth?

My understanding of P. vs NP is quite limited. I can understand P refers to an algorithm with an upper bound (big O) with order of growth $$n^c$$ for some constant c and variable n. My question is, do NP problems have a hypothesized lower bound order of growth (big Omega) of $$c^n$$ for deterministic machines? I can't find this stated anywhere and I'm trying to understand if this is something that is assumed or not.

Thanks.

The Exponential Time Hypothesis states that 3SAT requires exponential time $$c^m$$ for some constant $$c > 1$$, where $$m$$ is the number of variables. The Strong Exponential Time Hypothesis states that $$k$$SAT requires exponential time $$c_k^m$$, where $$c_k \to 2$$.
However, your conjecture is unconditionally false. Consider the language $$L$$ consisting of all pairs $$\langle x,y \rangle$$ such that $$x$$ is a 3SAT instance on $$m$$ variables (in particular, $$m \leq |x|$$), and $$y = 0^{|x|^{100}}$$. The language $$L$$ is clearly NP-complete, but it can be solved in time $$2^m \mathit{poly}(n) = 2^{n^{1/100}} \mathit{poly}(n)$$, where $$n$$ is the input size.
The Exponential Time Hypothesis does imply that every NP-complete problem requires time $$c^{n^\alpha}$$, for some $$c>1$$ and $$\alpha>0$$. This kind of running time is sometimes called subexponential time, but unfortunately this expression has several different interpretations, depending on context.