Your guess is correct. (The exponent should be $k+1$, even when the base is a point.) For any smooth fiber bundle
$$ F\hookrightarrow P \stackrel{\pi}{\to} M $$
we have a short exact sequence of vector bundles over $P$
$$ 0\to VTP\to TP \to \pi^* TM\to 0, $$
where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce
$$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$
This leads to your guess since the classical Euler exact sequence argument shows that
$$ c\bigl(\; V T\mathbb{P}(V)\;\bigr)= \bigl(\; 1+c_1(\gamma^*)\;\bigr)^{k+1}. $$
(Note the exponent is $k+1$.)