algebra precalculus – Is the distributive law of mutiplication valid when the number of terms are variable?

In brief the distributive law for natural numbers states that:

Let $a$,$b$,$c$ naturals then $a cdot (b+c) = acdot b + acdot c$

And this could be generalized by induction:

Let $a, a_1, a_2, ldots a_n$ naturals then $acdot (a_1 + a_2 + ldots +a_n) = a cdot a_1 + acdot a_2 + ldots +acdot a_n$

Those are valid propierties with other objects, for example, integers, real numbers, complex numbers, matrices, functions, etc. because their definition is based on this.

But what happens when $n$ is not a fixed natural but a variable? I’m interested in this because I want to know if the following is valid:

Let $N$ be a natural number, then:
$$ N = overbrace{1+1+ldots +1}^{N- times}$$ and multiplying by N on the left in both sides of the equality we get $$ Ncdot N = N cdot(overbrace{1+1+ldots +1}^{N- times}) $$ now the left hand side becomes $N^2$ and until this point I agree this is valid. But I’m not sure that $N cdot(overbrace{1+1+ldots +1}^{N- times}) = overbrace{N + N + ldots + N}^{N- times}$ becasue this will bring a contradiction if we think this as a function. $$N^2 = overbrace{N + N + ldots + N}^{N- times}$$ and derive both side with respect to N $$ frac{d}{dN}(N^2) = frac{d}{dN}left((overbrace{N + N + ldots + N}^{N- times})right) \
2N = overbrace{1 + 1 + ldots + 1}^{N-times} = N$$

And clearly the function $N$ is not equal to the function $2N$, hence a contradiction

I’m sure that the distributive law is failing with the derivative operator, but I’m not sure with the multiplication operator.

Being more specific:
if $c$ is a constant and $v$ a variable is the following assertion valid?
$$overbrace{ c + c +c ldots +c }^{v -text{times}} =^{?} c cdot(overbrace{1+1+cdot+1}^{ v-text{times}}) $$