Show that $y(t) = T(t − t_{0})$ also satisfies Newton’s law of cooling, $\frac{dT}{dt} = k(T_{e} − T)$, for any constant $t_{0}$.

Question

This question if from MIT's Open Course Ware for Single Variable Calculus:

Show that $\:y(t) = T(t − t_{0})$ also satisfies Newton’s law of cooling, $\: \frac{dT}{dt} = k(T_{e} - T,)\:$for any constant $t_{0}$. Write out the formula for $T(t − t_{0})$ and show that it is the same as the formula, $\:T = T_{e} + (T_{0} - T_{e})e^{-k(t - t_{0})}$(replacing t with $t - t_{0}$, this is the solution to the original differential equation given), for y(t) by identifying the constants k, $\:T_{e}$ and $\:T_{0}$ with their corresponding values in the formula.

This is the answer given:

To confirm the differential equation:

$\:y\prime(t) = T\prime(t - t_{0}) = k(T_{e} − T(t − t_{0})) = k(T_{e} − y(t))$

The formula for y is $y(t) = T(t − t_{0}) = Te + (T_{0} − T_{e})^{e−k(t−t_{0})} = a + (y(t_{0}) − a)^{e−c(t−t_{0})}$

with k = c, $T_{e}$ = a, and $T_{0}$ = $T(0)$ = $y(t_{0})$.

I don't understand why when y(t) is differentiated the product rule isn't applied and the derivative of the variable t is taken as well. When I attempted this first part, I found $\:y\prime(t) = \frac{dT}{dt}(t - t_{0}) + T.\:$ Then $\:y\prime(t) = k(T_{e} - T)(t-t_{0}) + T = k[(T_{e}(t - t_{0}) - T(t - t_{0})] + T = k((T_{e}(t - t_{0}) - y(t)) + T$

I don't understand how they found their $\:y\prime(t)$.

For the second part of their answer, I don't understand how $\:T_{0} = y(t_{0}).$ When you plug in $t_{0}$ you get $y(t_{0}) = T(t_{0} - t_{0}) = 0$.

Could someone please explain what I'm missing here?

Thank you so much!!!