Provided I have experimental couples $(x,y)$, is the adjustment of $\beta_i$ parameters of the following function:
$$y = f(x) = \beta_0 \cdot \exp (\beta_1\cdot x) + \beta_2$$
A Non Linear Least Squares problem because of the $\beta_2$ parameter?
I know that I can first estimate $\beta_2$ and then apply the following transformation:
$$y' = y - \beta_2$$
This will reduce the problem to a linear least square in term of parameters because the transformation $\log y'$ is possible.
I am just surprised that a simple offset factor could make a Least Square problem Non Linear. Am I right?
You are correct. The model being $$y = \beta_0 \, \exp (\beta_1\, x) + \beta_2$$ assuming that you know more or less the value of $\beta_2$, you can rewrite $$\log(y-\beta_2)=\log(\beta_0)+\beta_1x=\alpha+\beta_1\,x$$ and then linear regression will give you $\alpha$ and $\beta_1$ with $\beta_0=e^{\alpha}$.
But, take care : these are only estimates of the real values since, in this preliminary good step, you minimized $$SSQ_1=\sum_{i=1}^n\left(\alpha+\beta_1\,x_i-\log(y_i-\beta_2) \right)^2$$ which has little to do with $$SSQ_2=\sum_{i=1}^n\left(\beta_0 \, \exp (\beta_1\, x_i) + \beta_2-y_i\right)^2$$ You need to continue with the nonlinear regression.