Summary: We establish two truncations of Gauss' square exponent theorem and a finite extension of Euler's identity. For instance, we prove that for any positive integer $n$, $$ \sum _{k=0}^n(-1)^k \left [ \begin{matrix} 2n-k\\ k \end{matrix} \right ] (q;q^2)_{n-k}q^{{k+1\choose 2}} =\sum _{k=-n}^n(-1)^kq^{k^2}, $$ where $$ \left [ \begin{matrix} n\\ m\end{matrix} \right ] =\prod _{k=1}^m\frac {1-q^{n-k+1}}{1-q^k} \quad \text {and} \quad (a;q)_n=\prod _{k=0}^{n-1}(1-aq^k). $$
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