Ah ok, I misunderstood the question. So doing by hand is easy in principle. The basic idea of the Nyquist plot and the Bode plot, is identical. I know you only asks about the Nyquist plot, but the two are so similar, that it is good to consider both at the same time.
Remember that the transfer function is complex in value, and the input to the transfer function is the variable "s" , which is also complex. If you think about it, it is impossible to draw a plot of a complex function versus a complex input variable, because you would need 4 dimensions to plot it. However, if we restrict "s" to be imaginary only, with s=jw, where j=sqrt(-1), then we can make three dimensional plots of T(w) versus w. But, three dimensional plots are a little unwieldy, so typically we use two dimensional plots to keep it simple.
For the Bode plot, we make two separate two dimensional plots. One plot is the magnitude of the transfer function versus w, and the second is the phase of the transfer function versus w.
For the Nyquist plot, we make a 2d parametric plot of the Imaginary component of the transfer function versus the real part of the transfer function, and we let w be a parametric variable to describe the plot.
Mathematically, we can derive the Nyquist plot for your transfer function as follows.
T(s)=exp(-10 s)/(s+1)
T(jw)=exp( -10 j w)/(1+j w)=(cos(10 w) - j sin(10 w))/(1+jw)
T(w)=(cos(10 w) - j sin(10 w))*(1-jw)/(1+w^2)
Re(T(w))= (cos(10 w) - w sin(10 w))/(1+w^2)
Im(T(w))=(-sin(10 w)-w cos(10 w))/(1+w^2)
With the above two formulas, let w go from minus infinity to plus infinity and plot Im(T) versus Re(T) to get the parametric plot.
Please check my math, as I did this quickly on the fly to give you the general idea of how to do it.