= Solution
For <independent> Poisson counts, the <log-likelihood> at mean vector $\mu$ is
$$
\ell(\mu)=\sum_i\{y_i\log\mu_i-\mu_i-\log(y_i!)\}.
$$
The saturated model maximizes each cell separately at $\widehat\mu_i=y_i$, including the boundary value zero. If $e_i$ is the fitted mean under the log-linear regression, twice the saturated-minus-fitted <log-likelihood> is the full <Poisson deviance>
$$
D=2\sum_i\left\{y_i\log\frac{y_i}{e_i}-y_i+e_i\right\},
$$
with $0\log(0/e_i)=0$ understood by continuity. The fitted intercept has <score function> $\partial\ell/\partial\mu=\sum_i(y_i-e_i)$, since differentiating the logarithmic mean with respect to that intercept gives one. Its maximum-likelihood equation therefore gives $\sum_i e_i=\sum_i y_i$. This is why <Poisson deviance simplifies when an intercept is fitted>:
$$
\boxed{D=2\sum_i y_i\log(y_i/e_i)}.
$$
Without the fitted unpenalized intercept, one cannot generally discard the two linear terms. The individual logarithmic terms in the simplified sum need not each be positive; nonnegativity belongs to the full <deviance>, or to the full cell contributions before cancellation.
For the approximation, write $y_i=e_i+r_i$ and expand $\log(1+r_i/e_i)$. When relative residuals are small,
$$
\begin{aligned}
2\left\{(e_i+r_i)\log\left(1+\frac{r_i}{e_i}\right)-r_i\right\}
&=\frac{r_i^2}{e_i}-\frac{r_i^3}{3e_i^2}
+O\left(\frac{r_i^4}{e_i^3}\right).
\end{aligned}
$$
Consequently the <quadratic Pearson approximation to the Poisson deviance> is
$$
\boxed{D\simeq\sum_i\frac{(y_i-e_i)^2}{e_i}},
$$
the <Pearson chi-squared statistic> for unit Poisson dispersion. Adequate expected counts and small relative residuals are what make the quadratic approximation useful; it is not an exact identity and can be poor for sparse cells, especially a zero count with positive fitted mean. Any further chi-squared goodness-of-fit reference needs its own large-count regularity assumptions.
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