Look at the densities. By virtue of the definition of the Dirichlet distribution, on the right hand side the density (in variables $\mathbf x = (x_1,x_2,\ldots,x_k)$ with $x_1+x_2+\cdots+x_k=1$) is proportional to

$$F_\gamma(\mathbf x) = \frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}.$$

On the left hand side, because $\Gamma(\gamma_j+1)=\gamma_j\Gamma(\gamma_j),$ the density for term $j$ is proportional to

$$\gamma_j\frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}\frac{x_j}{\gamma_j} = x_jF_\gamma(\mathbf x ).$$

(The initial $\gamma_j$ comes from the coefficient in the linear combination -- ignoring the common factor of $1/\gamma$ -- while the final $\gamma_j$ in the denominator comes from the preceding $\Gamma$ identity.)

Summing over all $j$ produces $x_1+x_2+\cdots+x_k = 1$ times $F_\gamma(\mathbf x).$ Because both distributions necessarily are normalized to have unit integrals and are proportional to the same function of $\mathbf x$, they are equal, QED.

Look at the densities. By virtue of the definition of the Dirichlet distribution, on the right hand side the density (in variables $\mathbf x = (x_1,x_2,\ldots,x_k)$ with $x_1+x_2+\cdots+x_k=1$) is proportional to

$$F_\gamma(\mathbf x) = \frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}.$$

On the left hand side, because $\Gamma(\gamma_j+1)=\gamma_j\Gamma(\gamma_j),$ the density for term $j$ is proportional to

$$\gamma_j\frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}\frac{x_j}{\gamma_j} = x_jF_\gamma(\mathbf x ).$$

(The initial $\gamma_j$ comes from the coefficient $\beta_j$ in the linear combination -- ignoring the common factor of $1/\gamma$ -- while the final $\gamma_j$ in the denominator comes from the preceding $\Gamma$ identity.)

Summing over all $j$ produces $x_1+x_2+\cdots+x_k = 1$ times $F_\gamma(\mathbf x).$ Because both distributions necessarily are normalized to have unit integrals and are proportional to the same function of $\mathbf x$, they are equal, QED.