Let p be a prime, ε>0\documentclass[12pt]{minimal}
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\begin{document}$$\varepsilon >0$$\end{document} and 0<L+1<L+N<p\documentclass[12pt]{minimal}
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\begin{document}$$0<L+1<L+N < p$$\end{document}. We prove that if p1/2+ε<N<p1-ε\documentclass[12pt]{minimal}
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\begin{document}$$p^{1/2+\varepsilon }< N <p^{1-\varepsilon }$$\end{document}, then #{n!(modp);L+1≤n≤L+N}>c(NlogN)1/2,c=c(ε)>0.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \#\{n!\,\,({\mathrm{mod}} \,p);\,\, L+1\le n\le L+N\} > c (N\log N)^{1/2},\,\, c=c(\varepsilon )>0. \end{aligned}$$\end{document}We use this bound to show that any λ≢0(modp)\documentclass[12pt]{minimal}
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\begin{document}$$\lambda \not \equiv 0\ ({\mathrm{mod}}\, p)$$\end{document} can be represented in the form λ≡n1!⋯n7!(modp)\documentclass[12pt]{minimal}
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\begin{document}$$\lambda \equiv n_1!\cdots n_7!\ ({\mathrm{mod}}\, p)$$\end{document}, where ni=o(p11/12)\documentclass[12pt]{minimal}
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\begin{document}$$n_i=o(p^{11/12})$$\end{document}. This refines the previously known range for ni\documentclass[12pt]{minimal}
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\begin{document}$$n_i$$\end{document}.
