I'm trying to prove the symmetry of mixed partial derivatives in which the following lemma is appealed.
$\textbf{Lemma:}$ Let $X$ be open in $\mathbb R^n$, $f:X \to F$, and $m \in \mathbb N$. We $h \in \mathbb R$ and $j \in \{1,\ldots,n\}$, we define a map $\Delta_j^h f$ by $$\Delta_j^h f: X \to F, \quad x \mapsto f(x+he_j)-f(x)$$
Suppose $j_1, j_2, \ldots, j_m \in\{1,\ldots,n\}$ and $\sigma$ is a permutation of $\{1, \ldots, m\}$. Then $$ \Delta_{j_1}^{h_1} \Delta_{j_2}^{h_2} \cdots\Delta_{j_m}^{h_m} f (a) = \Delta_{j_{\sigma(1)}}^{h_{\sigma(1)}} \Delta_{j_{\sigma(2)}}^{h_{\sigma(2)}} \cdots\Delta_{j_{\sigma(m)}}^{h_{\sigma(m)}} f (a)$$
Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
$\textbf{My attempt:}$
Because every permutation is a product of adjacent transpositions, it suffices to prove that $$\Delta_{j_1}^{h_1} \Delta_{j_2}^{h_2} f (a) = \Delta_{j_2}^{h_2} \Delta_{j_1}^{h_1} f (a)$$
We have $$\begin{aligned} \Delta_{j_1}^{h_1} \Delta_{j_2}^{h_2} f (a)& = \Delta_{j_1}^{h_1} \left ( f(a+h_2 e_{j_2})-f(a) \right)\\ &= \Delta_{j_1}^{h_1} f(a+h_2 e_{j_2}) - \Delta_{j_1}^{h_1} f(a) \\ &= (f(a+h_2 e_{j_2} + h_1 e_{j_1})- f(a+h_2 e_{j_2})) - ( f(a+h_1 e_{j_1} ) - f(a)) \\ &= f(a+h_2 e_{j_2} + h_1 e_{j_1})- f(a+h_2 e_{j_2}) - f(a+h_1 e_{j_1} ) + f(a)\end{aligned}$$ and $$\begin{aligned} \Delta_{j_2}^{h_2} \Delta_{j_1}^{h_1} f (a)& = \Delta_{j_2}^{h_2} \left ( f(a+h_1 e_{j_1})-f(a) \right)\\ &= \Delta_{j_2}^{h_2} f(a+h_1 e_{j_1}) - \Delta_{j_2}^{h_2} f(a) \\ &= (f(a+ h_1 e_{j_1}+h_2 e_{j_2} )- f(a+h_1 e_{j_1})) - ( f(a+h_2 e_{j_2} ) - f(a)) \\ &= f(a + h_1 e_{j_1}+h_2 e_{j_2})- f(a+h_1 e_{j_1}) - f(a+h_2 e_{j_2} ) + f(a)\end{aligned}$$
Hence the expansions of $\Delta_{j_1}^{h_1} \Delta_{j_2}^{h_2} f (a)$ and $\Delta_{j_2}^{h_2} \Delta_{j_1}^{h_1} f (a)$ are identical. Thus $\Delta_{j_1}^{h_1} \Delta_{j_2}^{h_2} f (a) = \Delta_{j_2}^{h_2} \Delta_{j_1}^{h_1} f (a)$. This completes the proof.