Morera's theorem of entire function

For each fixed $n$, show that $$f_n(z)=\int_1^nt^{z-1}e^{-t}dt$$ is an
entire function of $z$.

From Morera ‘s theorem:

If a continuous, complex-valued function $f$ in a domain $D$ that
satisfies

$$\oint_\gamma f(z)\,dz = 0 $$

for every closed contour in $D$, then $f$ is analytic.

From this theorem how can I show that my function is indeed continuous? Do I prove this straight from the definition of continuity?

A hint which was provided, which I don’t understand why, goes:$$\text{Look at:} \ |f(z_1)-f(z_2)|\le \int_1^n\frac{e^{-t}}{t}|t^{z_1}-t^{z_2}|dt \ \text{where},\\ z_1=x_1+iy_1 \\ z_2=x_2+iy_2$$

thus I must bound this but I don’t know how to go from here? Any hints please?

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Related problems: I. Just see this,

$$ \oint_\gamma f_n(z) dz = \oint_\gamma \int_1^nt^{z-1}e^{-t}dt dz = \int_1^n e^{-t}\frac{dt}{t}\oint_\gamma e^{z\ln t} dz \,dt = \dots. $$

Can you finish it. What kind of function is $e^{z\ln(t)}$?

Added: To prove continuity, advance like that

$$ |f_n(z+h)-f_n(z)| = \Big|\int_1^n (e^{(z+h)\ln t}-e^{ z \ln t }) \frac{e^{-t}}{t}dt \Big| $$

$$ \leq \int_1^n \big|e^{(z+h)\ln t}-e^{ z \ln t }\big| \frac{e^{-t}}{t}dt \leq M e^{-1} |h|\int_{1}^{n} \ln t\frac{1}{t}dt\dots. $$

Can you finish the task? Notice that, I used the mean value theorem to find the last inequality and $ e^{-t} \leq e^{-1} $ for $t\in [1,n]$.