Discrete Fourier Transform and Opposite Convolution Theorem
I'm not sure I understand your argument, but here's a proof which should make the relationship clear:

$$IDFT\left\{\frac{1}{N}(X*Y_N)_k\right\}=\frac{1}{N^2}\sum_{k=0}^{N-1}(X*Y_N)_ke^{i2\pi nk/N}=\\
=\frac{1}{N^2}\sum_{k=0}^{N-1}\sum_{l=0}^{N-1}X_lY_{(k-l)_N}e^{i2\pi nk/N}=
\frac{1}{N^2}\sum_{l=0}^{N-1}X_l\sum_{k=0}^{N-1}Y_{(k-l)_N}e^{i2\pi nk/N}=\\
\frac{1}{N^2}\sum_{l=0}^{N-1}X_le^{i2\pi nl/N}\sum_{k=0}^{N-1}Y_{(k-l)_N}e^{i2\pi n(k-l)/N}=\\=\frac{1}{N}\sum_{l=0}^{N-1}X_le^{i2\pi nl/N}\cdot\frac{1}{N}\sum_{k=0}^{N-1}Y_{k}e^{i2\pi nk/N}=x_ny_n$$

where $(k-l)_N$ means $(k-l)$ modulo $N$. I hope this helps.