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fij belong in G not in the Gamma-quotients
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Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$, and define a series $\Gamma_i$ of subgroups of $G$ by letting $\Gamma_1 = G$ and $$ \Gamma_{i+1} = \langle [ \Gamma_i,G ], \Gamma ^p _{\lceil (i+1)/p \rceil} \rangle .$$ Then $\Gamma_i / \Gamma_{i+1}$ is elementary abelian, so we can fix elements $f_{i1}, \ldots, f_{id_i}$ of $G$ whose images in $\Gamma_i/\Gamma_{i+1}$ form a basis. Consider all products of the form

$$ \prod_{i,j} (f_{ij}-1)^{\alpha_{ij}} \in kG \qquad (1) $$ where the product is taken in lexicographic order and $0 \leqslant \alpha_{ij} \leqslant p-1$. Define the weight of such a product to be $\sum_{i,j} i\alpha_{ij}$.

Jennings' Theorem says that if $J=I(G)=\operatorname{rad}(kG)$ then the set of products (1) of weight at least $s$ form a basis of $J^s$, and a basis for $J^s/J^{s+1}$ is given by the images of the products of weight exactly $s$.

In particular, the largest non-zero power of the radical is $$ \sum i (p-1) \dim_{\mathbb{F}_p} (\Gamma_i/\Gamma_{i+1}) $$

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