Let $X_i$ be the result of $i$-th die. Thesis: $P\left(\sum_{i=1}^{N} X_i = k\right) \text{ is equal to coefficient in $z^k$ of } \left(\frac{z^1+\ldots +z^6}{6}\right)^N. \quad\quad\quad(1) $
Proof by induction:
Base of induction: $P(X_1 = k) = \frac{1}{6}$ for $1 \leq k \leq 6$ and $0$ otherwise.
Indeed this is the case for $\left(\frac{z^1+\ldots+z^6}{6}\right)^1$.
Assumption of induction: thesis (1) holds for $1, 2, \ldots, N$.
Hypothesis of induction: thesis (1) holds for $N+1$.
Step of induction:
We know that $ \left(\frac{z^1+\ldots+z^6}{6}\right)^{N+1} = \left(\frac{z^1+\ldots+z^6}{6}\right)^N\left(\frac{z^1+\ldots+z^6}{6}\right) \quad\quad\quad(2)$ By the assumption we know that first part of right-hand side represents the probability distribution of sum of $N$ dice and the second part represents single die. Using the notation I've used in another answer we can observe that right-hand side can be written as $W_{X_1 + \ldots + X_N} \cdot W_{X_{N+1}}$ and that equals $W_{X_1 + \ldots + X_N + X_{N+1}}$ by the formulas I have derived there. But definition of this polynomial is $ W_{X_1 + \ldots + X_N + X_{N+1}}(z) = \sum_k P(X_1 + \ldots + X_N + X_{N+1} = k) z^k $
so the coefficient at $z^k$ is $P(X_1 + \ldots + X_N + X_{N+1} = k)$ which is precisely the induction hypothesis and that completes the induction step.
By the method of induction that completes the proof of (1) for all $N \geq 1$.
Afterword: It is important that different dice get different $X$-es, because this way you say that those results are independent. Having just one $X$ would make possible to derive that you have only 6 possible answers: $N, 2N, 3N, \ldots, 6N$, as it would mean just taking the very same result of the single $die$ $N$ times. (Notation $W_{X+X}$ in one of the previous comments is just wrong and shouldn't have happened. That ought to be $W_{X_1 + X_2}$ naturally.)
In conclusion, I think I overdid it a little, but I guess more explicit is in this case better than less explicit, please bear it. Also there may be some typos there, so watch out!