Why is lithium the most reducing alkali metal, and not caesium?
The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated.

$$\begin{array}{cc}
\hline
\ce{M} &amp; E^\circ(\ce{M+}/\ce{M}) \\
\hline 
\ce{Li} &amp; -3.045 \\
\ce{Na} &amp; -2.714 \\
\ce{K} &amp; -2.925 \\
\ce{Rb} &amp; -2.925 \\
\ce{Cs} &amp; -2.923 \\
\hline
\end{array}$$
Source: Chemistry of the Elements 2nd ed., Greenwood &amp; Earnshaw, p 75

However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward.



The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction

$$\ce{M(s) -&gt; M+(aq) + e-}$$

by the equation

$$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$

where $K$ is the absolute standard Gibbs free energy for the reaction

$$\ce{H+ + e- -&gt; 1/2 H2}$$

and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on.

$$\require{AMScd}
\begin{CD}
\ce{M (s)} @&gt;{\large \Delta_\mathrm{r}H}&gt;&gt; \ce{M+(aq) + e-} \\
@V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\
\ce{M (g)} @&gt;&gt;{\large IE_1(\ce{M})}&gt; \ce{M+ (g) + e-}
\end{CD}$$

We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$:

$$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$

(the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone, this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$.

However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$.

$$\begin{array}{ccccc}
\hline
\ce{M} &amp; \Delta_\mathrm{atom}H &amp; IE_1 &amp; \Delta_\mathrm{hyd}H &amp; \text{Sum} \\
\hline
\ce{Li} &amp; 161 &amp; 520 &amp; \mathbf{-520} &amp; 161 \\
\ce{Cs} &amp; 79 &amp; 376 &amp; \mathbf{-264} &amp; 211 \\
\hline
\end{array}$$
Source: Inorganic Chemistry 6th ed., Shriver et al., p 160

This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) &gt; E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion.

Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood &amp; Earnshaw, p 75).