Let's define, for any $i=1,2,...,n$ $$ h_{i}=\left\vert a_{ii}\right\vert -\sum_{j=1,j\neq i}\left\vert a_{ij}\right\vert $$ Then, by strict diagonally dominance, one has $h_{i}>0$ $\forall i$ Let $D=diag\{\left( h_{1}\right) ^{-1},\left( h_{2}\right) ^{-1},...,\left( h_{n}\right) ^{-1}\}$ and $B=DA$ so that the i-th row of $B $ matrix is equal to the corresponding row of $A$ matrix multiplied by $% \left( h_{i}\right) ^{-1}$ In this way , one has \begin{eqnarray*} d_{i} &=&\left\vert b_{ii}\right\vert -\sum_{j=1,j\neq i}\left\vert b_{ij}\right\vert \\ &=&\frac{\left\vert a_{ii}\right\vert }{h_{i}}-\sum_{j=1,j\neq i}\frac{% \left\vert a_{ij}\right\vert }{h_{i}} \\ &=&1 \end{eqnarray*}Now, let $\lambda $ be an eigenvalue of $B$ and $v=[v_{1},v_{2},...,v_{n}]$ the corresponding eigenvector; let moreover $p$ be the index of the maximal component of $v$ i.e. $$ \left\vert v_{p}\right\vert \geq \left\vert v_{i}\right\vert \text{ \ }% \forall i $$ Of course, by definition of eigenvector, $\left\vert v_{p}\right\vert >0$ Writing the p-th characteristic equation, we have:\begin{eqnarray*} \lambda v_{p} &=&\sum_{j=1}^{n}b_{pj}v_{j} \\ &=&b_{pp}v_{p}+\sum_{j=1,j\neq p}^{n}b_{pj}v_{j} \end{eqnarray*}so that, being $\left\vert \frac{v_{j}}{v_{p}}\right\vert \leq 1$ \begin{eqnarray*} \lambda &=&b_{pp}+\sum_{j=1,j\neq p}^{n}b_{pj}\frac{v_{j}}{v_{p}} \\ \left\vert \lambda \right\vert &=&\left\vert b_{pp}+\sum_{j=1,j\neq p}^{n}b_{pj}\frac{v_{j}}{v_{p}}\right\vert \\ &\geq &\left\vert \left\vert b_{pp}\right\vert -\left\vert \sum_{j=1,j\neq p}^{n}b_{pj}\frac{v_{j}}{v_{p}}\right\vert \right\vert \\ &\geq &\left\vert \left\vert b_{pp}\right\vert -\sum_{j=1,j\neq p}^{n}\left\vert b_{pj}\right\vert \left\vert \frac{v_{j}}{v_{p}}\right\vert \right\vert \text{ \ \ \ \ \ \ \ \ \ (*)} \\ \\ &\geq &\left\vert \left\vert b_{pp}\right\vert -\sum_{j=1,j\neq p}^{n}\left\vert b_{pj}\right\vert \right\vert \text{ \ \ \ \ \ \ \ \ \ (**)} \\ \\ &=&\left\vert b_{pp}\right\vert -\sum_{j=1,j\neq p}^{n}\left\vert b_{pj}\right\vert \\ &=&d_{p}=1 \end{eqnarray*}In this way, we found that each eigenvalue of $B$ is greater than one in absolute value; for this reason,$$ \left\vert \det (B)\right\vert =\left\vert \prod_{i=1}^{n}\lambda _{i}\right\vert \geq 1 $$ Finally,$$ \det (D)=\prod_{i=1}^{n}\left( h_{i}\right) ^{-1}=\left( \prod_{i=1}^{n}h_{i}\right) ^{-1} $$ so that \begin{eqnarray*} 1 &\leq &\left\vert \det (B)\right\vert \\ &=&\left\vert \det (D)\right\vert \left\vert \det (A)\right\vert \\ &=&\left( \prod_{i=1}^{n}h_{i}\right) ^{-1}\left\vert \det (A)\right\vert \end{eqnarray*}whence the thesis.

Remark: Perhaps it could be not immediately evident where the hypothesis of strict diagonally dominance is employed in this proof; in fact, inequality (*) and (**) would be, in a general case, not valid; they can be stated only because we can assure, by virtue of strict diagonally dominance, that the final argument of the absolute value ($% \left\vert b_{pp}\right\vert -\sum_{j=1,j\neq p}^{n}\left\vert b_{pj}\right\vert $ does remain positive.