Proof: The number $N$ of graphs with degrees $d_1,\ldots,d_n$ is the coefficient of $x_1^{d_1}\cdots x_n^{d_n}$ in the generating function $\prod_{j\lt k}(1+x_jx_k)$. Now apply Cauchy's Theorem in $n$ complex dimensions to find that $$N = \frac{1}{(2\pi i)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+x_jx_k)}{x_1^{d_1+1}\cdots x_n^{d_n+1}} dx_1\cdots dx_n,$$ where each integral is a simple closed contour enclosing the origin once. Choosing the circles $x_j=e^{i\theta_j}$, we get $$N = \frac{1}{(2\pi)^n} \oint\cdots\oint int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{d_1\theta_1+\cdots k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n}} d_n\theta_n)}} d\theta_1\cdots d\theta_n.$$ Alternatively, choosing the circles $x_j=e^{i(\theta_j+\pi)}$, we get $$N = \frac{1}{(2\pi)^n} \oint\cdots\oint int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{d_1\theta_1+\cdots k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n+k\pi}} d_n\theta_n+k\pi)}} d\theta_1\cdots d\theta_n,$$ where $k=d_1+\cdots+d_n$. Since $e^{ik\pi}=-1$ when $k$ is an odd integer, we can add these two integrals to get $2N=0$.
Proof: The number $N$ of graphs with degrees $d_1,\ldots,d_n$ is the coefficient of $x_1^{d_1}\cdots x_n^{d_n}$ in the generating function $\prod_{j\lt k}(1+x_jx_k)$. Now apply Cauchy's Theorem in $n$ complex dimensions to find that $$N = \frac{1}{(2\pi i)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+x_jx_k)}{x_1^{d_1+1}\cdots x_n^{d_n+1}} dx_1\cdots dx_n,$$ where each integral is a simple closed contour enclosing the origin once. Choosing the circles $x_j=e^{i\theta_j}$, we get $$N = \frac{1}{(2\pi)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{d_1\theta_1+\cdots +d_n\theta_n}} d\theta_1\cdots d\theta_n.$$ Alternatively, choosing the circles $x_j=e^{i(\theta_j+\pi)}$, we get $$N = \frac{1}{(2\pi)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{d_1\theta_1+\cdots +d_n\theta_n+k\pi}} d\theta_1\cdots d\theta_n,$$ where $k=d_1+\cdots+d_n$. Since $e^{ik\pi}=-1$ when $k$ is an odd integer, we can add these two integrals to get $2N=0$.