The proof uses three types of elementary operations (each achievable by multiplication by an invertible matrix):

1. Swap two rows (or columns).
2. Multiply a row (or column) by a unit.
3. Add a multiple of one row (or column) to another.

Step 1: If $A = 0$, we are done. Otherwise, by swapping rows/columns, we may assume $a_{11} \neq 0$.

Step 2: Make $a_{11}$ divide all entries in the first row and column. If $a_{11} \nmid a_{1j}$, perform the Euclidean algorithm: write $a_{1j} = a_{11} q + r$ with $0 \neq r$ and $\varphi(r) < \varphi(a_{11})$ (where $\varphi$ is the Euclidean function; in a general PID, use that ideals are principal to argue). Replace column $j$ by column $j - q \cdot$ column $1$, getting $r$ in position $(1,j)$. Swap columns to put $r$ in position $(1,1)$. The "size" of $a_{11}$ strictly decreases, so this terminates. Similarly for column entries.

Step 3: After Step 2, $a_{11}$ divides all entries in row 1 and column 1. Subtract multiples to zero them out: column $j \leftarrow$ column $j - (a_{1j}/a_{11}) \cdot$ column 1 for $j \geq 2$, and similarly for rows.

Step 4: If $a_{11}$ does not divide some $a_{ij}$ with $i, j \geq 2$: add row $i$ to row 1, creating a nonzero entry $a_{1j} = a_{ij}$ in row 1 not divisible by $a_{11}$. Return to Step 2. Since the ideal $(a_{11})$ strictly increases each time this happens, and $R$ is Noetherian, this terminates.

Step 5: Now $a_{11}$ divides all entries of $A$. The matrix has the form $\begin{pmatrix} a_{11} & 0 \\ 0 & A' \end{pmatrix}$. Apply the procedure recursively to $A'$.

Step 6: The resulting diagonal entries $d_1, \ldots, d_k$ satisfy $d_i \mid d_{i+1}$ by construction (Step 4 ensures each diagonal entry divides all later ones). $\blacksquare$